1. The class mark of classes 90-120 is :
(A) 90 (B) 105 (C) 115 (D) 120
1. The class mark of classes 90-120 is :
(A) 90 (B) 105 (C) 115 (D) 120
Explanation:
The upper class in the provided class of 90–120 is 120, and the lower class is 90.
Having said that,
Upper-class mark plus Lower class 2
=120+90/2=210/2=105.
2. The range of the data:25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is (A) 10 (B) 15 (C) 18 (D) 26
2. The range of the data:25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is (A) 10 (B) 15 (C) 18 (D) 26
Explanation:
The right response is D 26.
Determine the data set's range.
Justification for the ideal selection:
Choice (A). 26
25,18,20,22,16,6,17,15,12,30,32,10,19,8,11,20 are the numbers provided.
The data's maximum value, in this case, is 32.
The data's smallest value is six.
Now, Range=maximum value-minimum
value=32-6=26
The range, therefore, is 26.
So, D is the right answer.
Justification for bad choices:
since the data's range is 26. Therefore, options (A), (B), and (C) are erroneous.
So, D is the right answer.
3. In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is : (A) 6 (B) 7 (C) 8 (D) 12
3. In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is : (A) 6 (B) 7 (C) 8 (D) 12
Explanation:
The right response is B 7.
explanation of the ideal choice
The interval's width is equal to 6 and the class's median is 10
The formula Lowerlimit=Midvalue-Width2
=10-6/2
=10-3
=7 is known.
Therefore, Option B is the right choice.
4. The width of each of the five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. The upper-class limit of the highest class is: (A) 15 (B) 25 (C) 35 (D) 40
4. The width of each of the five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. The upper-class limit of the highest class is: (A) 15 (B) 25 (C) 35 (D) 40
Explanation:
The right response is C 35.
When the lower class limit of the lowest class in a frequency distribution is 10 and each of the five continuous classes has a width of 5, the upper-class limit of the frequency distribution can be calculated as follows:
The maximum first-class price is 10 + 5 = 15.
The fifth-class upper limit is 35 (15 + 4 x 5).
5. Let m be the mid-point and l be the upper-class limit of a class in a continuous frequency distribution. The lower class limit of the class is : (A) 2m + l (B) 2m – l (C) m – l (D) m – 2l
5. Let m be the mid-point and l be the upper-class limit of a class in a continuous frequency distribution. The lower class limit of the class is : (A) 2m + l (B) 2m – l (C) m – l (D) m – 2l
Explanation:
Let x and y represent a continuous frequency distribution's bottom and upper-class limits.
Now, given the class midpoint of
=x+y2=m.
⇒x+y=2m⇒x+l=2m
[∵y = l = provided upper-class limit]
⇒x=2m−l
Therefore, the class's lower class limit is 2m–l.
6. The class marks of a frequency distribution are given as follows: 15, 20, 25, ... The class corresponding to the class mark 20 is : (A) 12.5 – 17.5 (B) 17.5 – 22.5 (C) 18.5 – 21.5 (D) 19.5 – 20.5
6. The class marks of a frequency distribution are given as follows: 15, 20, 25, ... The class corresponding to the class mark 20 is : (A) 12.5 – 17.5 (B) 17.5 – 22.5 (C) 18.5 – 21.5 (D) 19.5 – 20.5
Explanation:
The response is B.
5 cents separates the midpoint from the range.
As a result, the class that corresponds to the class mark 20 must have a difference of 5.
Option (c) and (d) are therefore incorrect.
Because the midpoint is 20, we can only choose option (b).
∵17.522.5/2
=402
=20.
7. In the class intervals 10-20, and 20-30, the number 20 is included in (A) 10-20 (B) 20-30 (C) both the intervals (D) none of these intervals
7. In the class intervals 10-20, and 20-30, the number 20 is included in (A) 10-20 (B) 20-30 (C) both the intervals (D) none of these intervals
Explanation:
The right answer is B) The number is always included in the class interval's lower limit. 20 is therefore included in 2030.
8. A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data: 268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236. The frequency of the class 310-330 is: (A) 4 (B) 5 (C) 6 (D) 7
8. A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data: 268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236. The frequency of the class 310-330 is: (A) 4 (B) 5 (C) 6 (D) 7
Explanation:
The correct answer is C) 310,310,320,319,318,316 are the numbers between 310 and 330. This interval will not include the number 330. Frequency is therefore 6.
9. A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the classes is constructed for the following data: 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44. The number of classes in the distribution will be : (A) 9 (B) 10 (C) 11 (D) 12
9. A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the classes is constructed for the following data: 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44. The number of classes in the distribution will be : (A) 9 (B) 10 (C) 11 (D) 12
Explanation:
The right response is B 10.
The frequency range is between 14 and 112.
The class intervals are 13-22
23-32
33-42
43-52
for the given set of data.
53-62
63-72
73-82
83-92
93-102
103-112
The number of class intervals is therefore 10.
10. To draw a histogram to represent the following frequency distribution :
Class interval 5-10 10-15 15-25 25-45 45-75 Frequency 6 12 10 8 15 The adjusted frequency for the class 25-45 is : (A) 6 (B) 5 (C) 3 (D) 2
10. To draw a histogram to represent the following frequency distribution :
Class interval 5-10 10-15 15-25 25-45 45-75 Frequency 6 12 10 8 15 The adjusted frequency for the class 25-45 is : (A) 6 (B) 5 (C) 3 (D) 2
Explanation:
The class intervals are not the same size in this case. Therefore, we use the following formula to determine the adjusted frequency: Adjustedfreq=minclasssizeclasssizeofthisclassfreq
Class sizes are 10 – 5 = 5.
15 - 10 = 5
25 - 15 = 10
45 - 25 = 20
75 - 45 = 30
The minimum class size is five.
11. The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is : (A) 28 (B) 30 (C) 35 (D) 38
11. The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is : (A) 28 (B) 30 (C) 35 (D) 38
Explanation:
The solution is D.
We'll use x1, x2, x3, x4, and x5.
The average of the five digits is 30, thus.
⇒x1+x2+x3+x4+x55=30
⇒x1+x2+x3+x4+x5=150
⇒x1+x2+x3+x4=150−x5
The result of multiplying both sides by 4 is x1+x2+x3+x45=150x54.(i)
Given, the four numbers' means are 28 150 x 54 = 28 [from Eq. (i)].
⇒150−x5=112
⇒x5=150−112
⇒x5=38
38 is hence the excluded number.
12. If the mean of the observations: x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is (a)10 ⅓(b)10 ⅔ (c)11 ⅓ (d) 11 ⅔
12. If the mean of the observations: x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is (a)10 ⅓(b)10 ⅔ (c)11 ⅓ (d) 11 ⅔
Explanation:
The response is C.
In light of this, the observations x,x+3,x+5,x+7, and x+10 have a mean of 9
x+x+3+x+5+x+7+x+10/5 =9
⇒5x+25=45
⇒5x=20
⇒x=4
The final three findings are x+5=4+5=9 and x+7=4+7=11.
And x+10=4+10=14
So, the average of the most recent three observations is 9+11+14/3
=34/3=11 ⅓
The average of the most recent three measurements is therefore 11 ⅓.
13. If x represents the mean of n observations x1 , x2, ..., xn , then the value of (x− x)is:(A) –1 (B) 0 (C) 1 (D) n – 1 (A) –1 (B) 0 (C) 1 (D) n – 1
13. If x represents the mean of n observations x1 , x2, ..., xn , then the value of (x− x)is:(A) –1 (B) 0 (C) 1 (D) n – 1 (A) –1 (B) 0 (C) 1 (D) n – 1
Explanation:
B) is the right answer because xˉ is the average of all the observations. Half of the numbers will be on the right side of the mean and the other half will be on the left if we plot all the numbers together with the mean on a number line. Because of this, (xi−xˉ) will produce an equal number of positive and negative words, whose sum is 0.
14. If each observation of the data is increased by 5, then their mean (A) remains the same (B) becomes 5 times the original mean (C) is decreased by 5 (D) is increased by 5
14. If each observation of the data is increased by 5, then their mean (A) remains the same (B) becomes 5 times the original mean (C) is decreased by 5 (D) is increased by 5
Explanation:
Take into account x1, x2, x3, x4,... xn as the n observations Old mean xold=∑ni=1/n ==1. (1)
When each observation is multiplied by 5, the new mean is given as x new = (x1 + 5) + (x2 + 5) +... + (xn + 5)/n.
The formula is xnew = (x1 + x2 +.... + xn) + 5n/n.
Consequently, we have x new = ∑ni=1xi/n∑=1 + 5 (by utilizing equation (1)).
x new = x old plus 5
Consequently, their mean has risen by 5 points.
15. Let x be the mean of x1 , x2 , ... , xn and y the mean of y1 , y2 , ... , yn . If z is the mean of x1 , x2 , ... , xn , y1 , y2 , ... , yn , then z is equal to (A) x y + (B)x y / 2 (C) x +y/n(d)x +y/2n
15. Let x be the mean of x1 , x2 , ... , xn and y the mean of y1 , y2 , ... , yn . If z is the mean of x1 , x2 , ... , xn , y1 , y2 , ... , yn , then z is equal to (A) x y + (B)x y / 2 (C) x +y/n(d)x +y/2n
Explanation:
The right answer is (b): 12( x+ y)
Given: ¯¯¯x=(x1+x2+.....xn)/n ⇒x1+x2+......+xn=n¯¯¯x……(i)
Additionally, y=(y1+y2+......+yn)/n y1+y2+......+yn=n y......(ii) At this point, z=(x1+x2+...+xn)+(y1+y2+...+yn)/2n z=(n¯¯¯x+n¯¯¯y)2n From equations (i) and (ii),
=12(¯¯¯x+¯¯¯y)
16. If x is the mean of x1 , x2 , ... , xn , then for a ≠ 0, the mean of ax1 , ax2 , ..., axn x1/a x2/a…….,xn/a is
A (a+1/a) x ()b (a+1/a) (c)(a+1/a) x/n (d) (a+1/a) x/2n
16. If x is the mean of x1 , x2 , ... , xn , then for a ≠ 0, the mean of ax1 , ax2 , ..., axn x1/a x2/a…….,xn/a is
A (a+1/a) x ()b (a+1/a) (c)(a+1/a) x/n (d) (a+1/a) x/2n
Explanation:
RESPONSE: (b) (a+1/a)/x.
Required mean is 1/2(ax +1/ax ̄)(ax1+ax2+...+axn)+(ax ̄1/ax2/a+...+axn/a)/2n.
17. If 1x , 2 x , 3 x , ... , n x are the means of n groups with n1 , n2 , ... , n number of observations respectively, then the mean x of all the groups taken together is given by :
17. If 1x , 2 x , 3 x , ... , n x are the means of n groups with n1 , n2 , ... , n number of observations respectively, then the mean x of all the groups taken together is given by :
Explanation:
(c) Ni=1 Ni Xi=1 Ni
Sum of all terms equals (n1x ̄1)+(n2x ̄2)+(nnx ̄n)
Terms: (n1 + n2 +... nn)
∴Mean=(∑ni=1ni¯xi∑ni=1ni
18. The mean of 100 observations is 50. If one of the observations which were 50 is replaced by 150, the resulting mean will be : (A) 50.5 (B) 51 (C) 51.5 (D) 52
18. The mean of 100 observations is 50. If one of the observations which were 50 is replaced by 150, the resulting mean will be : (A) 50.5 (B) 51 (C) 51.5 (D) 52
Explanation:
(b) 51
100 observations at a mean of 50
The sum of 100 observations equals 100 x 50, or 5000.
Given is the replacement of one of the observations, 50, with 150.
∴ (5000 - 50 + 150)/new sum = 5100
And,
the resultant mean is 5100100=51.
19. There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be –3.5. The mean of the given numbers is : (A) 46.5 (B) 49.5 (C) 53.5 (D) 56.5
19. There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be –3.5. The mean of the given numbers is : (A) 46.5 (B) 49.5 (C) 53.5 (D) 56.5
Explanation:
The right answer is D) Total numbers equal 50.
Numbers after removing 53 from each have a mean of 3.5 and a total of 3.5 times 50, which equals 175
The total of the initial numbers is 175+5350=2825.
50/2825, the original figures' median, equals 56.5.
20. The mean of 25 observations is 36. Out of these observations if the mean of the first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation (A) 23 (B) 36 (C) 38 (D) 40
20. The mean of 25 observations is 36. Out of these observations if the mean of the first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation (A) 23 (B) 36 (C) 38 (D) 40
Explanation:
Idea: 2 Marks
2 Marks for Application
The first 13 observations' mean value is 32.
The sum of the first 13 observations is equal to 416= 32x 13
The latest 13 observations' mean is 39.
The last 13 observations' sum equals (39x13)=507
25 observations on average equal 36.
The sum of all 25 observations is equal to 900 (36 x 25).
The thirteenth observation is equal to (416 + 507 - 900) = 23.
The 13th observation is therefore 23.
21. The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is (A) 45 (B) 49.5 (C) 54 (D) 56
21. The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is (A) 45 (B) 49.5 (C) 54 (D) 56
Explanation:
Let's sort the given observations 22, 34, 39, 45, 54, 54, 56, 68, 78, and 84 in ascending order.
n = 10 observations overall
Let's use the median formula since n is even.
The median is equal to [(n/2)nd observation plus (n/2 + 1)th observation]/2.
The value = [(10/2)th observation + (10/2 + 1)th observation]/2 = [5th observation + 6th observation]/2 is substituted.
Thus, we obtain = [54 + 54]/2 = 108/2 = 54.
As a result, 54 is the data's median.
22. For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and whose abcissae are respectively : (A) upper limits of the classes (B) lower limits of the classes (C) class marks of the classes (D) upper limits of perceeding classes
22. For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and whose abcissae are respectively : (A) upper limits of the classes (B) lower limits of the classes (C) class marks of the classes (D) upper limits of perceeding classes
Explanation:
Frequency It is possible to describe polygons as a type of graph that interprets data or information that is frequently utilized in statistics.
The distance from the origin along the x-axis at the point (x, y) is indicated by the abscissa, which is the value of x in the point.
Ordinate is the y value at the coordinates (x, y) and the angle of the point with respect to the x-axis.
In a frequency distribution, a class mark is the midpoint or middle value of a certain class.
As a result, the abscissae represent the respective class marks.
23. Median of the following numbers: 4, 4, 5, 7, 6, 7, 7, 12, 3 is (A) 4 (B) 5 (C) 6 (D) 7
23. Median of the following numbers: 4, 4, 5, 7, 6, 7, 7, 12, 3 is (A) 4 (B) 5 (C) 6 (D) 7
Explanation:
Let's put the numbers in the following order: 3, 4, 5, 6, 7, 7, 12, n = 9.
Since n in this instance is odd, the calculation for the median is
[(n + 1)/2]th observation is the median.
changing the value
[(9 + 1)/2] is the median. It was observed
(10)/2th observation is the median
Thus, we obtain Median = Fifth Observation.
Average = 6
Consequently, six is the median.
24. Mode of the data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15 is (A) 14 (B) 15 (C) 16 (D) 17
24. Mode of the data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15 is (A) 14 (B) 15 (C) 16 (D) 17
Explanation:
The response is B.
The data is first arranged as follows in ascending order.
14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 20
The most frequent number, 15, is shown above, occurring 5 times.
Consequently, the given data's mode is 15.
25. In a sample study of 642 people, it was found that 514 people have a high school certificate. If a person is selected at random, the probability that the person has a high school certificate is : (A) 0.5 (B) 0.6 (C) 0.7 (D) 0.8
25. In a sample study of 642 people, it was found that 514 people have a high school certificate. If a person is selected at random, the probability that the person has a high school certificate is : (A) 0.5 (B) 0.6 (C) 0.7 (D) 0.8
Explanation:
The total number of participants in the sample study (n(S) = 642) is known.
The number of individuals with a high school diploma, n(E), is 514.
We are aware that P = n(E)/n(S) is the probability that a person has a high school diploma.
514/642 divided by the values equals 0.8.
Therefore, there is a 0.8 chance that they have a high school diploma.
26. In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he/she does not like to eat potato chips is : (A) 0.25 (B) 0.50 (C) 0.75 (D) 0.80
26. In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he/she does not like to eat potato chips is : (A) 0.25 (B) 0.50 (C) 0.75 (D) 0.80
Explanation:
364 youngsters in all were surveyed.
There were 91 kids that liked eating potato chips.
Number of kids who didn't like potato chips = 364 91 = 273 P(Child doesn't like to eat potato chips) = Kids who didn't enjoy potato chips/Total kids surveyed=273_/364=34
Option (c) is the proper response as a result.
27. In a medical examination of students of a class, the following blood groups are recorded: Blood group A AB B O Number of students 10 13 12 5
A student is selected at random from the class. The probability that he/she has blood group B, is: A ¼ B 13/40 c 3/10 d ⅛
27. In a medical examination of students of a class, the following blood groups are recorded: Blood group A AB B O Number of students 10 13 12 5
A student is selected at random from the class. The probability that he/she has blood group B, is: A ¼ B 13/40 c 3/10 d ⅛
Explanation:
The response is C.
n(S) = 40 is the total number of students taking the medical examination.
Number of people with blood type B, n(E) = 12 Blood group B likelihood = n(E) /n(S) = 12/40 = 3/10.
28. Two coins are tossed 1000 times and the outcomes are recorded as below: Number of heads 2 1 0 Frequency 200 550 250 Based on this information, the probability for at most one head is (A) ⅕ (B)¼ © ⅘ (D)3/4
28. Two coins are tossed 1000 times and the outcomes are recorded as below: Number of heads 2 1 0 Frequency 200 550 250 Based on this information, the probability for at most one head is (A) ⅕ (B)¼ © ⅘ (D)3/4
Explanation:
The response is C.
n(S) = 1000 is the total number of coins tossed.
N = 550 + 250 = 800 is the number of outcomes with at most one head. The probability for a maximum of one head is equal to n(E)/n(S)=800/1000=45.
Therefore, the likelihood of at least one head is 45.
29. 80 bulbs are selected at random from a lot and their lifetime (in hrs) is recorded in the form of a frequency table given below :
Lifetime (in hours) 300 500 700 900 1100 Frequency 10 12 23 25 10 One bulb is selected at random from the lot. The probability that its life is 1150 hours, is (A)1/80 (b)7/16 © 0 (D) 1
29. 80 bulbs are selected at random from a lot and their lifetime (in hrs) is recorded in the form of a frequency table given below :
Lifetime (in hours) 300 500 700 900 1100 Frequency 10 12 23 25 10 One bulb is selected at random from the lot. The probability that its life is 1150 hours, is (A)1/80 (b)7/16 © 0 (D) 1
Explanation:
The response is C.
n(S) = 80 is the number of bulbs in the lot.
There are 1150 bulbs in existence, hence n(E) = 0. The likelihood that it will last 1150 hours is n(E)/n(S)=0/80=0.
Consequently, there is no chance that it will live past 1150.
30. Refer to Q.29 above The probability that bulbs selected randomly from the lot has life less than 900 hours is :
(A)11/40 (B)5/16 © 7/16 (D)9/16
30. Refer to Q.29 above The probability that bulbs selected randomly from the lot has life less than 900 hours is :
(A)11/40 (B)5/16 © 7/16 (D)9/16
Explanation:
The solution is D.
n(S) = 80 is the total number of bulbs in the batch.
n = 10 + 12 + 23 = 45, the number of bulbs with a life expectancy of fewer than 900 hours.
∴ The probability that a bulb will last less than 900 hours is equal to n(E)/n(S)=45/80=9/16.
31. The frequency distribution :
Marks 0-20 20-40 40-60 60-100
Number of Students 10 15 20 25
has been represented graphically
Do you think this representation is correct? Why?
31. The frequency distribution :
Marks 0-20 20-40 40-60 60-100
Number of Students 10 15 20 25
has been represented graphically
Do you think this representation is correct? Why?
Explanation:
The right answer is A) According to the graph, 10 students received 0–20 marks.
15 pupils earned 20/40 points.
Twenty pupils received 40–60 and twenty-five received 60–100.
The frequency distribution table demonstrates the same.
The appropriate histogram, complete with changing widths, is shown below.
32. In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded: 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98, 44 Which ‘average’ will be a good representative of the above data and why?
32. In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded: 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98, 44 Which ‘average’ will be a good representative of the above data and why?
Explanation:
Solution
Find the most accurate average for the provided data:
The data are 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98, and 44.
It is obvious that every value only appears once.
11,40,41,44,46,48,52,53,54,62,96,98 are the rearranged data.
Thus, excessive values have an impact on the data.
As a result, the Median can serve as the finest representation of the supplied data's average.
As a result, the median can serve as the best representation of the average.
33. A child says that the median of 3, 14, 18, 20, and 5 is 18. What doesn’t the child understand about finding the median?
33. A child says that the median of 3, 14, 18, 20, and 5 is 18. What doesn’t the child understand about finding the median?
Explanation:
We are aware that the median denotes the value in the middle of any group. At this stage, half of the data is more and half is less.
The median makes it possible to express a lot of data points with just one. The median is the most straightforward statistical metric to compute.
The middle data point reflects the median of the data after the data are organized in ascending order for the purpose of calculating the median.
Given that, 3, 14, 18, 20, and 5 have a median of 18,
The data must be sorted in ascending order before being used to calculate the median.
The child has not sorted the provided information in ascending order.
The child does not comprehend that information as a result.
As a result, the child is unaware that information must be put in ascending sequence.
34. A football player scored the following number of goals in the 10 matches: 1, 3, 2, 5, 8, 6, 1, 4, 7, 9 Since the number of matches is 10 (an even number), therefore, the median = 5th besrvation+6th obeservation/2
=8+6/2=7 is its the correct answer and why?
34. A football player scored the following number of goals in the 10 matches: 1, 3, 2, 5, 8, 6, 1, 4, 7, 9 Since the number of matches is 10 (an even number), therefore, the median = 5th besrvation+6th obeservation/2
=8+6/2=7 is its the correct answer and why?
Explanation:
The numbers provided are 1, 3, 2, 5, 8, 6, 14, 7, and 9.
Putting it in this order: 1, 1, 2, 3, 4, 5, 6, 7, 8, 9
10 observations are made, which is an even number.
(n/2)nd observation plus (n/2 + 1)nd observation divided by two.
Changing the values
The median is equal to the product of the second and the tenth observations.
Median = (4 + 5)/2 Median = (5th observation + 6th observation)/2
Therefore, we obtain Median = 9/2 Median = 4.5.
Therefore, since the data must be organized in either ascending or descending order before computing the median, solution 7 is incorrect.
35. Is it correct to say that in a histogram, the area of each rectangle is proportional to the class size of the corresponding class interval? If not, correct the statement.
35. Is it correct to say that in a histogram, the area of each rectangle is proportional to the class size of the corresponding class interval? If not, correct the statement.
Explanation:
Verify the information provided.
A histogram depicts fluctuating data along the vertical axis while the horizontal axis represents class intervals with fixed widths. As a result, the width of each rectangle in the histogram remains constant but the length varies. The area of each rectangle in a histogram is therefore proportional to the corresponding frequency of its class.
Consequently, the assertion is untrue.
36. The class marks of a continuous distribution are: 1.04, 1.14, 1.24, 1.34, 1.44, 1.54 and 1.64 Is it correct to say that the last interval will be 1.55 - 1.73? Justify your answer.
36. The class marks of a continuous distribution are: 1.04, 1.14, 1.24, 1.34, 1.44, 1.54 and 1.64 Is it correct to say that the last interval will be 1.55 - 1.73? Justify your answer.
Explanation:
It is inaccurate. because there should be a class size proportional to the difference between two consecutive class grades.
Here, the class size ranges from 1.55 to 1.73, which is not equal, and the difference between two consecutive marks is 0.1.
37. 30 children were asked about the number of hours they watched TV programs last week. The results are recorded as under :
Number of hours 0-5 5-10 10-15 15-20 Frequency 8 16 4 2
Can we say that the number of children who watched TV for 10 or more hours a week is 22? Justify your answer.
37. 30 children were asked about the number of hours they watched TV programs last week. The results are recorded as under :
Number of hours 0-5 5-10 10-15 15-20 Frequency 8 16 4 2
Can we say that the number of children who watched TV for 10 or more hours a week is 22? Justify your answer.
Explanation:
The number of kids who watched TV for 10 or more hours is 10-15 and 15-20 hours in the accompanying table.
Thus, we may format it as
Children who watched television for 10 or more hours totaled 10-15 and 15-20 hours, which equals 4 + 2 = 6.
Therefore, it is untrue to state that there were 22 kids who watched TV for 10 or more hours per week.
Therefore, we cannot claim that 22 kids watched TV for ten or more hours every week.
38. Can the experimental probability of an event be a negative number? If not, why?
38. Can the experimental probability of an event be a negative number? If not, why?
Explanation:
Not at all.
The total number of trials is always positive and the number of trials in which the event can occur cannot be negative.
39. Can the experimental probability of an event be greater than 1? Justify your answer.
39. Can the experimental probability of an event be greater than 1? Justify your answer.
Explanation:
We are aware that probability equals the number of trials an event occurs in. Total trials conducted
The maximum number of trials in which an event could occur is the total number of trials.
Therefore, an event's experimental probability cannot be higher than 1.
40. As the number of tosses of a coin increase, the ratio of the number of heads to the total number of tosses will be ½ . Is it correct? If not, write the correct one.
40. As the number of tosses of a coin increase, the ratio of the number of heads to the total number of tosses will be ½ . Is it correct? If not, write the correct one.
Explanation:
The ratio of heads to total tosses will increase with the number of coin tosses and will eventually approach but not quite reach 1/2.
The claim is untrue as a result.
41. The blood groups of 30 students are recorded as follows:
A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB, B,
A, A, O, A, AB, B, A, O, B, A, B, A
Prepare a frequency distribution table for the data.
41. The blood groups of 30 students are recorded as follows:
A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB, B,
A, A, O, A, AB, B, A, O, B, A, B, A
Prepare a frequency distribution table for the data.
Explanation:
As a result, the blood type of the class's 30 students can be represented as follows.
42. The value of π upto 35 decimal places is given below:
3. 14159265358979323846264338327950288
Make a frequency distribution of the digits 0 to 9 after the decimal point.
42. The value of π upto 35 decimal places is given below:
3. 14159265358979323846264338327950288
Make a frequency distribution of the digits 0 to 9 after the decimal point.
Explanation:
The number of observations inside a certain interval is shown through a representation called a frequency distribution, which can be either graphical or tabular in nature.
The frequency distribution table is now created:
Digits Frequency
0 1
1 2
2 5
3 6
4 3
5 4
6 3
7 2
8 5
9 4
43. The scores (out of 100) obtained by 33 students in a mathematics test are as follows:
69, 48, 84, 58, 48, 73, 83, 48, 66, 58, 84000
66, 64, 71, 64, 66, 69, 66, 83, 66, 69, 71
81, 71, 73, 69, 66, 66, 64, 58, 64, 69, 69
Represent this data in the form of a frequency distribution.
43. The scores (out of 100) obtained by 33 students in a mathematics test are as follows:
69, 48, 84, 58, 48, 73, 83, 48, 66, 58, 84000
66, 64, 71, 64, 66, 69, 66, 83, 66, 69, 71
81, 71, 73, 69, 66, 66, 64, 58, 64, 69, 69
Represent this data in the form of a frequency distribution.
Explanation:
The frequency of the mark is the percentage of pupils who receive the same marks in mathematics.
The data's frequency distribution table is as follows:
The frequency distribution of the provided data is so described above.
44. Prepare a continuous grouped frequency distribution from the following data:
Mid-point Frequency
5 4
15 8
25 13
35 12
45 6
44. Prepare a continuous grouped frequency distribution from the following data:
Mid-point Frequency
5 4
15 8
25 13
35 12
45 6
Explanation:
15-5, or 10 is the difference between the two midpoints.
It shows that the class interval is 10 wide.
If an is the lower limit of the first class interval, then a + 10 is the upper limit.
First-class interval at mid-class = 5
The mid value is calculated as (Lower limit + Upper limit)/2.
Changing the values
5 = (a + a + 10)/2
Additional math shows that 5 = (2a + 10)/2.
2a + 10 = 10
Thus, we obtain 2a = 0
a = 0.
The first period is from 0 to 10.
The table showing the continuous grouped frequency distribution is
Consequently, the class interval is 10 in size.
45. Convert the given frequency distribution into a continuous grouped frequency distribution:
Class interval Frequency
150-153 7
154-157 7
158-161 15
162-165 10
166-169 5
170-173 6
In which intervals would 153.5 and 157.5 be included?
45. Convert the given frequency distribution into a continuous grouped frequency distribution:
Class interval Frequency
150-153 7
154-157 7
158-161 15
162-165 10
166-169 5
170-173 6
In which intervals would 153.5 and 157.5 be included?
Explanation:
Given, the frequency distribution is shown in the table.
We need to determine the interval in which 153.5 and 157.5 falls.
The frequency distribution offered is inclusive in nature.
Now that you've converted to exclusive form, think about classes 150–153 and 154–157.
154 is the lower bound of the range 154-157.
The range 150–153 has a maximum of 153.
(154 - 153/2)/2 = 0.5 is half the difference between the lower and higher limits.
As a result, we add 0.5 to each upper boundary and 0.5 to each lower limit.
The continuous grouped frequency distribution table looks like this:
Using the table above,
We note that the intervals between 153.5 and 157.5, respectively, are 153.5 - 157.5 and 157.5 - 161.5.
46. The expenditure of a family on different heads in a month is given below:
Head Food Education Clothing House Rent Others Savings
Expenditure 4000 2500 1000 3500 2500 1500
(in Rs)
Draw a bar graph to represent the data above.
46. The expenditure of a family on different heads in a month is given below:
Head Food Education Clothing House Rent Others Savings
Expenditure 4000 2500 1000 3500 2500 1500
(in Rs)
Draw a bar graph to represent the data above.
Explanation:
The steps that follow involve creating a bar graph using this data.
Step I: Since the width of the bar is unimportant, we represent the head (variable) on the horizontal axis using any scale. But for the sake of clarity, we keep the equal spacing between each bar and assume identical widths for all of them. Let one unit stand in for one head.
Step 2: On the vertical axis, we display the expense. The scale can be set to 1 unit = Rs. 500 because the maximum spending is Rs. 4000.
Step III: We create a rectangular bar with a width of 1 unit and a height of 8 units to represent our first head, which is food.
Step IV: In a similar manner, other heads are represented by spacing out two successive bars by 12 units.
Below is a bar graph of the information provided.
47. Expenditure on Education of a country during a five-year period (2002-2006), in crores of rupees, is given below:
Elementary education 240
Secondary Education 120
University Education 190
Teacher’s Training 20
Social Education 10
Other Educational Programmes 115
Cultural programs 25
Technical Education 125
Represent the information above with a bar graph.
47. Expenditure on Education of a country during a five-year period (2002-2006), in crores of rupees, is given below:
Elementary education 240
Secondary Education 120
University Education 190
Teacher’s Training 20
Social Education 10
Other Educational Programmes 115
Cultural programs 25
Technical Education 125
Represent the information above with a bar graph.
Explanation:
Given, the table shows a country's education spending over a five-year period (2002–2006), expressed in crores of rupees.
We must use a bar graph to illustrate the data.
Any scale can be used to show a nation's education on the horizontal axis.
Let one unit stand in for one head.
The expense can be shown on the vertical axis.
Because the whole cost is Rs. 240 crore
Let's say that one unit equals 25 crore rupees.
Think of the first head, which is elementary schooling.
The amount spent is Rs. 240 crores.
as 1 unit equals 25 crores rupees
9.6 units = 240 billion rupees
We create a rectangle bar that is 1 unit wide and 9.6 units tall.
Similar to how other heads are drawn, a gap is left.
48. The following table gives the frequencies of the most commonly used letters a, e, i, o, r, t, u from a page of a book :
Letters a e i o r t u
Frequency 75 125 80 70 80 95 75
Represent the information above with a bar graph.
48. The following table gives the frequencies of the most commonly used letters a, e, i, o, r, t, u from a page of a book :
Letters a e i o r t u
Frequency 75 125 80 70 80 95 75
Represent the information above with a bar graph.
Explanation:
The steps that follow involve creating a bar graph using this data.
Step I: We use any scale to display the letters (the variable) on the horizontal axis because the width of the bar is irrelevant. For the sake of clarity, we keep the equal spacing between each bar while using equal widths for each. Let each unit correspond to one letter.
The letters are represented on the vertical axis in step two. Since 125 is the highest possible frequency, we can choose a scale where 1 unit equals 15 frequencies.
Step III: Draw a rectangular bar with a width of one unit and a height of five units to symbolize our first letter.
Step IV - In a similar manner, other heads are represented by spacing out two successive bars by 12 units.
Below is a bar graph of the information provided.
49. If the mean of the following data is 20.2, find the value of p:
x 10 15 20 25 30
f 6 8 p 10 6
49. If the mean of the following data is 20.2, find the value of p:
x 10 15 20 25 30
f 6 8 p 10 6
Explanation:
Given that the data's median is 20.2
We must ascertain what p is worth.
Ungrouped data mean = Mean = ∑fx /∑ f
In where f is x's frequencies
Mean = [(10 × 6) + (15 × 8) + (20 × p) + (25 × 10) + (30 × 6)] / (6 + 8 + p + 10 + 6)
So, 20.2 = [60 + 120 + 20p + 250 +180] / (14 +p +16)
20.2 = [180 + 180 + 250 + 20p] / (30 + p)
20.2(30 + p) = 360 + 250 + 20p 606 + 20.2p = 610 + 20p
If you rearrange, 20.2p - 20p equals 610 - 606
0.2p = 4 p = 4/0.2 p = 20
P thus has a value of 20.
50. Obtain the mean of the following distribution:
Frequency Variable
4 4
8 6
14 8
11 10
3 12
50. Obtain the mean of the following distribution:
Frequency Variable
4 4
8 6
14 8
11 10
3 12
Explanation:
Given, the frequency distribution is shown in the table.
We must ascertain the distribution's mean.
Ungrouped data mean = Mean = ∑fx / ∑ f
In where f is x's frequencies
Mean = [(4 × 4) + (8 × 6) + (14 × 8) + (11 × 10) + (3 × 12)] / (4 + 8 + 14 + 11 + 3) = [16 + 48 + 112 + 110 + 36] / (12 + 14 + 14) = (42 + 48 + 222) / (12 + 28) = (90 + 222) / (40) = 322 / 40 = 8.05
Consequently, 8.05 is the distribution's mean.
51. A class consists of 50 students out of which 30 are girls. The mean marks scored by girls in a test is 73 (out of 100) and that of boys is 71. Determine the mean score of the whole class.
51. A class consists of 50 students out of which 30 are girls. The mean marks scored by girls in a test is 73 (out of 100) and that of boys is 71. Determine the mean score of the whole class.
Explanation:
Given that there are 50 students in the class, 30 of them are female.
The average test result for females is 73 out of 100.
The average test result for boys is 71 out of 100.
The average grade for the entire class must be found.
Mean is equal to the total of all observations divided by the total number of observations.
There are 30 girls.
50 boys less 30 boys equal 20 boys.
The female average score equals all female marks divided by the total female population
73 = total girls' marks divided by 30 total girls' marks equals 73(30) = 2190
Mean score for boys is calculated as follows: boys x total marks/boys
71 = total boys' marks divided by 20 total girls' marks
boys alone = 71(20) = 1420
Mean of the class = (total of all girls' marks plus total of all boys' marks) / (total of all girls' and boys' marks) = (2190 + 1420) / (30 + 20) = 3610/50 = 361/5 = 72.2
Consequently, the average for the entire class is 72.2.
52. Mean of 50 observations was found to be 80.4. But later on, it was discovered that 96 was misread as 69 in one place. Find the correct mean.
52. Mean of 50 observations was found to be 80.4. But later on, it was discovered that 96 was misread as 69 in one place. Find the correct mean.
Explanation:
The average of 50 observations is 80.4, so
It was revealed that 96 was sometimes read incorrectly as 69.
We must determine the proper mean.
Mean is equal to the total of all observations divided by the total number of observations.
Considering, mean = 80.4
50 observations were made.
Total observations added up = 80.4(50)
= 804(5)
= 4020.
Consequently, the correct mean is= (4020 - 69 + 96)/50
= (4116 - 69)/50
= 4047/50
= 80.94.
Thus, 80.94 is the proper mean.
53. Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. The median of the data is 24. Find the value of x.
53. Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. The median of the data is 24. Find the value of x.
Explanation:
Ten observations are given, and they are listed in ascending order.
6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, and 43 are the observations.
The data's median is 24.
The value of x must be determined.
There were n = 10 observations in total.
Since n is even,
Given by is the formula for the median when n is even.
Median is equal to [(n/2)nd observation plus (n/2)+1st observation] / 2
In other words, the median is equal to [(10/2)th observation + ((10/2) + 1)th observation] / 2 24 = [5th observation + (5 + 1)th observation] / 2
24 = [5th observation + 6th observation] / 2
24 = [(x + 1) + (2x - 13)]/2
48 = x + 2x + 1 - 13
48 = 3x - 12
3x = 48 + 12
3x = 60
x = 60/3
x = 20
Consequently, x has a value of 20.
54. The points scored by a basketball team in a series of matches are as follows: 17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28 Find the median and mode for the data.
54. The points scored by a basketball team in a series of matches are as follows: 17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28 Find the median and mode for the data.
Explanation:
Given that, a basketball team has scored 17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, and 28 points in a series of matches.
We must determine the data's median and mode.
In descending order, the points are 2, 5, 7, 7, 8, 10, 10, 10, 14, 17, 18, 24, 25, 27, 28, and 48.
There were 16 observations in all.
Since n is even,
Given by is the formula for the median when n is even.
Median is equal to [(n/2)nd observation plus (n/2)+1st observation] / 2
Therefore, median is equal to [(16/2)nd observation plus ((16/2) + 1)th observation] / 2 = [8th observation plus (8 + 1)th observation] / 2
= [8th observation plus 9th observation] / 2
= (10 + 14)/2
= 24/2
= 12
Consequently, 12 is the median.
The term "mode" refers to the observation that is made as often as possible
As we can see, 10 is repeated three times.
Consequently, mode is 10.
55. In Fig. 14.2, there is a histogram depicting the daily wages of workers in a factory. Construct the frequency distribution table.
55. In Fig. 14.2, there is a histogram depicting the daily wages of workers in a factory. Construct the frequency distribution table.
Explanation:
Given that the daily pay of factory workers is represented by the histogram.
A frequency distribution table must be created.
We can see from the histogram that the class interval has a width of 50 and is comprised of the ranges 150-200, 200-250, etc.
The table of frequency distribution is provided below.
The total number of employees is equal to 145 (50 + 35 + 40 + 20 + 10).
56. A company selected 4000 households at random and surveyed them to find out a relationship between income level and the number of television sets in a home. The information so obtained is listed in the following table:
Monthly income Number of Televisions/household
(in Rs) 0 1 2 Above 2
< 10000 20 80 10 0
10000 - 14999 10 240 60 0
15000 - 19999 0 380 120 30
20000 - 24999 0 520 370 80
25000 and above 0 1100 760 220
Find the probability:
(i) of a household earning Rs 10000 – Rs 14999 per year and having exactly one television.
(ii) of a household earning Rs 25000 and more per year and owning 2 televisions.
(iii) of a household not having any television.
56. A company selected 4000 households at random and surveyed them to find out a relationship between income level and the number of television sets in a home. The information so obtained is listed in the following table:
Monthly income Number of Televisions/household
(in Rs) 0 1 2 Above 2
< 10000 20 80 10 0
10000 - 14999 10 240 60 0
15000 - 19999 0 380 120 30
20000 - 24999 0 520 370 80
25000 and above 0 1100 760 220
Find the probability:
(i) of a household earning Rs 10000 – Rs 14999 per year and having exactly one television.
(ii) of a household earning Rs 25000 and more per year and owning 2 televisions.
(iii) of a household not having any television.
Explanation:
Given, the table illustrates the connection between household television ownership and income level.
A business randomly picked 4000 households to survey.
Given, the table illustrates the connection between household television ownership and income level.
A business randomly picked 4000 households to survey.
The likelihood of a household possessing exactly one television and earning between Rs 10,000 and Rs 14,999 per year must be calculated.
the likelihood of an event is Trials in which the occurrence has occurred divided by the total number of trials
According to the provided table, there are 240 households with exactly one television and annual incomes between Rs. 10,000 and Rs. 14,999.
Thus, there have been 240 trials in which the event has occurred.
Trails in total: 4000
Probability of choosing a home with exactly one television and an annual income of between Rs. 10,000 and Rs. 14,999 = 240/4000
= 24/400
= 12/200
= 6/100
= 0.06
Consequently, 0.06 is the necessary probability.
57. Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:
Sum Frequency
2 14
3 30
4 42
5 55
6 72
7 75
8 70
9 53
10 46
11 28
12 15
If the dice are thrown once more, what is the probability of getting a sum
(i) 3? (ii) more than 10? (iii) less than or equal to 5? (iv) between 8 and 12?
57. Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:
Sum Frequency
2 14
3 30
4 42
5 55
6 72
7 75
8 70
9 53
10 46
11 28
12 15
If the dice are thrown once more, what is the probability of getting a sum
(i) 3? (ii) more than 10? (iii) less than or equal to 5? (iv) between 8 and 12?
Explanation:
Given, 500 occurrences of simultaneous two-dice throwing will occur.
Each time, the two numbers that appear on their tops are added together and noted.
The frequency and sum of the numbers on the top of the dice are shown in the table.
The likelihood that the dice will be thrown again with a result between 8 and 12 must be calculated.
An event's probability is determined by dividing the number of trials in which it occurred by the total number of trials.
Using the provided table,
When we get a sum between 8 and 12 times, we get: 53 + 46 + 28
= 99 + 28
= 127.
Therefore, there have been 127 trials in which the event has occurred.
There are 500 trails overall.
The likelihood of receiving an amount between 8 and 12 is 127/500.
= 0.254
Consequently, 0.254 is the necessary probability.
58. Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table:
Number of defective bulbs 0 1 23456 more than 6
Frequency 400 180 48 41 18 8 3 2
One carton was selected at random. What is the probability that it has
(i) No defective bulb?
(ii) defective bulbs from 2 to 6?
(iii) defective bulbs less than 4?
58. Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table:
Number of defective bulbs 0 1 23456 more than 6
Frequency 400 180 48 41 18 8 3 2
One carton was selected at random. What is the probability that it has
(i) No defective bulb?
(ii) defective bulbs from 2 to 6?
(iii) defective bulbs less than 4?
Explanation:
Given that bulbs are packaged in boxes with 40 bulbs each
We checked 700 packages for broken bulbs.
The likelihood of choosing a carton with 2 to 6 flawed bulbs must be determined.
An event's probability is determined by dividing the number of trials in which it occurred by the total number of trials.
According to the data, there is 118 cartons total that have two to six faulty bulbs.
Therefore, there have been 118 trials in which the event has occurred.
There are 700 trails in all.
The probability of choosing a box with between 2 and 6 faulty bulbs is 118/700 59/350.
As a result, the likelihood that there are 2 to 6 damaged bulbs is = 59/350.
59. Over the past 200 working days, the number of defective parts produced by a machine is given in the following table:
Number of 0123456789 10 11 12 13
defective parts
Days 50 32 22 18 12 12 10 10 10 86622
Determine the probability that tomorrow’s output will have
(i) No defective part
(ii) at least one defective part
(iii) not more than 5 defective parts
(iv) more than 13 defective parts
59. Over the past 200 working days, the number of defective parts produced by a machine is given in the following table:
Number of 0123456789 10 11 12 13
defective parts
Days 50 32 22 18 12 12 10 10 10 86622
Determine the probability that tomorrow’s output will have
(i) No defective part
(ii) at least one defective part
(iii) not more than 5 defective parts
(iv) more than 13 defective parts
Explanation:
Given, the table shows how many defective parts a machine created during the course of 200 working days.
We need to calculate the likelihood that at least one defective component will be included in tomorrow's output.
An event's probability is determined by dividing the number of trials in which it occurred by the total number of trials.
The number of days with at least one defective item from the supplied table is 50 + 32 + 22 + 18 + 12 + 12 + 10 + 10 + 10 + 8 + 6 + 6 + 2 + 2 = 150.
Thus, there have been 150 trials in which the event has occurred.
There were 200 trials in total.
The probability that at least one defective component will be present in tomorrow's output is 150/200
= 75/100
= 3/4
= 0.75
Consequently, the likelihood that at least one damaged component will be present in tomorrow's output.
60. A recent survey found that the ages of workers in a factory are distributed as follows:
Age (in years) 20 - 29 30 - 39 40 - 49 50 - 59 60 and above
Number of workers 38 27 86 46 3
If a person is selected at random, find the probability that the person is:
(i) 40 years or more
(ii) under 40 years
(iii) having age from 30 to 39 years
(iv) under 60 but over 39 years
60. A recent survey found that the ages of workers in a factory are distributed as follows:
Age (in years) 20 - 29 30 - 39 40 - 49 50 - 59 60 and above
Number of workers 38 27 86 46 3
If a person is selected at random, find the probability that the person is:
(i) 40 years or more
(ii) under 40 years
(iii) having age from 30 to 39 years
(iv) under 60 but over 39 years
Explanation:
Given that the age distribution of factory workers is shown in the table,
A name is chosen at random.
The likelihood that the person is 40 years or older must be determined.
An event's probability is determined by dividing the number of trials in which it occurred by the total number of trials.
According to the provided statistics, there are 135 employees who are 40 years of age or older (86 + 46 + 3).
Therefore, the event occurred in 135 trials.
200 workers totaled by 38 + 27 + 86 + 46 + 3
The likelihood that the person is 40 or older is 135/200, or
=27/40
= 0.675.
Therefore, there is a 0.675 percent chance that the person is over 40.
61. The following are the marks (out of 100) of 60 students in mathematics.
16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28,72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63,25, 36, 54, 44, 47, 27, 72, 17, 4, 30. Construct a grouped frequency distribution table with width 10 of each class starting from 0 - 9.
61. The following are the marks (out of 100) of 60 students in mathematics.
16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28,72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63,25, 36, 54, 44, 47, 27, 72, 17, 4, 30. Construct a grouped frequency distribution table with width 10 of each class starting from 0 - 9.
Explanation:
Arranging the following data in ascending order: 4, 5, 7, 7, 13, 15, 16, 16, 17, 17, 19, 24, 25, 26, 27, 28, 30, 31, 34, 34, 35, 35, 36, 36, 36, 42, 44, 45, 47, 48, 51, 52, 52, 54, 55, 55, 56, 56, 61, 62, 62, 63, 68, 70, 72, 72, 72, 74, 75, 75, 78, 80, 81, 85, 86, 86, 92, 95, 97.
60 total frequency.
62. Refer to Q1 above. Construct a grouped frequency distribution table with a width of 10 for each class, in such a way that one of the classes is 10 - 20 (20 not included).
62. Refer to Q1 above. Construct a grouped frequency distribution table with a width of 10 for each class, in such a way that one of the classes is 10 - 20 (20 not included).
Explanation:
We divide the provided information into groups such as 0-10, 10-20, 20-30, etc. Whereas the upper-class limit is not a member of that class. Each case has a class width of 10.
The following table includes the data's frequency distribution.
63. Draw a histogram of the following distribution :
Heights (in cm) Number of students
150 - 153 7
153 - 156 8
156 - 159 14
159 - 162 10
162 - 165 6
165 - 168 5
63. Draw a histogram of the following distribution :
Heights (in cm) Number of students
150 - 153 7
153 - 156 8
156 - 159 14
159 - 162 10
162 - 165 6
165 - 168 5
Explanation:
The frequency distribution supplied is obviously in the exclusive form.
We choose a suitable scale to express the class intervals of heights along the horizontal axis. A suitable scale is used to indicate the corresponding frequencies of the number of pupils along the vertical axis.
The offered ranges begin with 150 to 153. It denotes that there is a break () close to the origin, indicating that the graph is created with a scale that starts at 150.
Below is a histogram of the given distribution.
64. Draw a histogram to represent the following grouped frequency distribution :
Ages (in years) Number of teachers
20 - 24 10
25 - 29 28
30 - 34 32
35 - 39 48
40 - 44 50
45 - 49 35
50 - 54 12
64. Draw a histogram to represent the following grouped frequency distribution :
Ages (in years) Number of teachers
20 - 24 10
25 - 29 28
30 - 34 32
35 - 39 48
40 - 44 50
45 - 49 35
50 - 54 12
Explanation:
The frequency distribution offered is inclusive in nature. We first change it into a unique form.
Now think about the 20–24 and 25–29 age groups.
25 is the lowest limit of 25-29.
The maximum range of 20 to 24 is 24.
Therefore, 25242=12=0.5 is the difference's half.
Therefore, we add 0.5 to each upper boundary and 0.5 to each lower limit.
Below is a table with the continuous grouped frequency distribution.
The information provided thus takes on an exclusive shape.
We choose a suitable scale to represent the age class intervals along the horizontal axis. A suitable scale is used to indicate the corresponding frequencies of the number of teachers along the vertical axis.
Since the specified intervals range from 19.5 to 24.5, there is a break around the origin, indicating that the graph is drawn with a scale that starts at 19.5.
Below is a histogram of the given distribution.
65. The lengths of 62 leaves of a plant are measured in millimeters and the data is represented in the following table :
Length (in mm) Number of leaves
118 - 126 8
127 - 135 10
136 - 144 12
145 - 153 17
154 - 162 7
163 - 171 5
172 - 180 3
Draw a histogram to represent the data above.
65. The lengths of 62 leaves of a plant are measured in millimeters and the data is represented in the following table :
Length (in mm) Number of leaves
118 - 126 8
127 - 135 10
136 - 144 12
145 - 153 17
154 - 162 7
163 - 171 5
172 - 180 3
Draw a histogram to represent the data above.
Explanation:
Calculate the frequency distribution in step one.
Given that 62 plant leaves measured in millimeters in length
The table shows the number of leaves and their length.
For the provided data, a histogram must be drawn.
The frequency distribution offered is inclusive in nature.
When switching to the exclusive form,
Think about the classes 118-126 and 127-135 now.
127 is the lower bound of the range 127–135.
126 is the top limit of 118-126.
(127-126)2=0.5, which is equal to the difference between the lower and higher limits.
As a result, we add 0.5 to each upper boundary and 0.5 to each lower limit.
Create the table for the continuous grouped frequency distribution in step two.
The continuous grouped frequency distribution table looks like this:
As a result, the offered information takes on an exclusive shape.
Step 3: Plot the histogram for the data in the provided table.
We represent the class intervals of length on an appropriate scale along the horizontal axis.
On a suitable scale, the relevant frequencies of the number of leaves are shown along the Y-axis.
The intervals presented begin with 117.5-126.5.
It denotes that there is a break () near the origin, indicating that the graph was produced with a scale that starts at 117.5.
As a result, the preceding description includes a histogram for the length class interval and the related frequencies of the number of leaves on a plant with 62 leaves.
66. The marks obtained (out of 100) by a class of 80 students are given below :
Marks Number of students
10 - 20 6
20 - 30 17
30 - 50 15
50 - 70 16
70 - 100 26
Construct a histogram to represent the data above.
66. The marks obtained (out of 100) by a class of 80 students are given below :
Marks Number of students
10 - 20 6
20 - 30 17
30 - 50 15
50 - 70 16
70 - 100 26
Construct a histogram to represent the data above.
Explanation:
Given that the table shows the frequency distribution of the overall grades earned by an 80-student class.
For the aforementioned distribution, a histogram must be created.
We note that the frequency distribution's class intervals do not have uniform widths.
To make the areas of the rectangles in the histogram proportionate to the frequencies, we must change the lengths of the rectangles in the histogram.
The minimum class size divided by the class size of that class gives the adjusted frequency of a class.
20 divided by 10 is 10, which is the minimum class size.
Create a histogram now using the information below.
The age class intervals can be visualized along the horizontal axis.
Along the vertical axis, the matching frequencies of the number of teachers are shown.
Given that the range is 10 to 20,
This suggests that a break is a present close to the origin.
A histogram based on the provided data is shown below.
67. Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a highway:
Class interval (km/h) Frequency
30 - 40 3
40 - 50 6
50 - 60 25
60 - 70 65
70 - 80 50
80 - 90 28
90 - 100 14
Draw a histogram and frequency polygon representing the data above.
67. Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a highway:
Class interval (km/h) Frequency
30 - 40 3
40 - 50 6
50 - 60 25
60 - 70 65
70 - 80 50
80 - 90 28
90 - 100 14
Draw a histogram and frequency polygon representing the data above.
Explanation:
The frequency distribution shown is obviously an exclusive form.
We represent the class intervals of length on an appropriate scale along the horizontal axis.
On a suitable scale, the corresponding frequencies are shown along the vertical axis.
Class intervals serve as the bases for the rectangles, and the corresponding frequencies serve as the heights.
Let's create a histogram for this data and label B, C, D, E, F, G, and H, respectively, as the midpoints of the top of the rectangles. Here, the first and last classes range in size from 30 to 40 to 90 to 100.
Think about the hypothetical classes 20–30 and 100–110, each with a frequency of 0. These classes' respective class marks at locations A and l are 25 and 105.
68. Refer to Q. 7: Draw the frequency polygon representing the above data without drawing the histogram.
68. Refer to Q. 7: Draw the frequency polygon representing the above data without drawing the histogram.
Explanation:
Without a histogram, a frequency polygon must be drawn.
First, we determine the class grades for the courses, which are 30-40, 40-50, 50-60, and 60-70. The grade is equal to 30+402=702=35.
The class marks for the other classes can also be found in a similar manner.
The table for class grades is seen below.
By graphing the class grades along the horizontal axis and the frequency along the vertical axis, we may create a frequency polygon.
Plotting the points B(35, 3), C(45, 6), D(55, 25), E(65, 65), F(75, 50), G(85, 28), and H(95, 14) is now complete.
Plot a point with a frequency of 0 that corresponds to the classes 20 to 30 and 100 to 110 that you are examining. To obtain the desired frequency polygon, join all of these spots.
69. Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them.
Section A Section B
Marks Frequency Marks Frequency
0 - 15 5 0 - 15 3
15 - 30 12 15 - 30 16
30 - 45 28 30 - 45 25
45 - 60 30 45 - 60 27
60 - 75 35 60 - 75 40
75 - 90 13 75 - 90 10
Represent the marks of the students of both sections on the same graph by two frequency polygons. What do you observe?
69. Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them.
Section A Section B
Marks Frequency Marks Frequency
0 - 15 5 0 - 15 3
15 - 30 12 15 - 30 16
30 - 45 28 30 - 45 25
45 - 60 30 45 - 60 27
60 - 75 35 60 - 75 40
75 - 90 13 75 - 90 10
Represent the marks of the students of both sections on the same graph by two frequency polygons. What do you observe?
Explanation:
According to the grades they earned, the table shows the frequency distribution of students in sections A and B of a class.
Two frequency polygons must be used to show the student's grades from both portions on the same graph.
histogram.
To begin drawing a frequency polygon, we must first locate the mid marks.
Midpoints equal (0 + 15)/2 (15/2 = 7.5)
The class grades might be shown along the horizontal axis.
The vertical axis displays the matching frequencies.
The points are A(7.5, 5), B(22.5, 12), C(37.5, 28), D(52.5, 30), E(67.5, 35), and F(82.5, 13) for section A.
The points are H(7.5, 3), I(22.5, 16), J(37.5, 25), K(52.5, 27), L(67.5, 40), and M(82.5, 10), for section B.
Create a graph using the points and connect them all.
40 students in section B received the maximum score of 67.5, as shown by the frequency polygon of the two parts.
70. The mean of the following distribution is 50.
x f
10 17
30 5a + 3
50 32
70 7a – 11
90 19
Find the value of a and hence the frequencies of 30 and 70.
70. The mean of the following distribution is 50.
x f
10 17
30 5a + 3
50 32
70 7a – 11
90 19
Find the value of a and hence the frequencies of 30 and 70.
Explanation:
40 students in section B received the maximum score of 67.5, as shown by the frequency polygon of the two parts.
Given that the distribution below has a mean of 50.
We need to determine the frequency of 30 and 70 as well as the value of a.
Having said that, Ungrouped data mean = Mean = fx / f
In where f is x's frequencies
The result is that the mean is [(17 10) + [(5a + 3) 30] + (32 50] + [(7a - 11) 70] + (19 90)] / (17 + 5a + 3 + 32 + 7a - 11 + 19)
50 = [170 + 150a + 90 + 1600 + 490a - 770 + 1710] / (12a + 60)
50 = (3570 - 770 + 640a)/(12a + 60)
50(12a + 60) = 2800 + 640a
600 a + 3000 = 2800 + 640a
40a = 200
a = 200/40
a = 20/4
a = 5
640a - 600a = 3000 - 2800
Consequently, a has a value of 5.
The formula for 30 is 30 = 5a + 3
= 5(5) + 3
= 25 + 3
= 28.
70 = 7a - 11
=7( 5) - 11
=35 - 11
= 24
As a result, 28 and 24 are the frequencies of 30 and 70.
71. The mean marks (out of 100) of boys and girls in an examination are 70 and 73, respectively. If the mean mark of all the students in that examination is 71, find the ratio of the number of boys to the number of girls.
71. The mean marks (out of 100) of boys and girls in an examination are 70 and 73, respectively. If the mean mark of all the students in that examination is 71, find the ratio of the number of boys to the number of girls.
Explanation:
If there are x boys and y girls, then let x be the number of boys.
Boys score, on average, 70 points.
Mean marks for boys are calculated as total marks for boys divided by the total number of boys, or 70 = total marks for boys divided by x.
Mean grades for girls are calculated as follows: 73 = Total grades for girls/y, or Total grades for girls = 73y.
Mean grades for all students are calculated as follows: 71 = (70x+73y)/(x+y) or (total boys' marks + total girls' marks)/(total boys' marks + total girls' marks). Cross-multiply
71x+71y=70x+73y x = 2y
71(x+y) = 70x+73y
x=2y
It indicates that there are 2:1 more boys than girls.
Response: 2:1.
72. A total of 25 patients admitted to a hospital are tested for levels of blood sugar, (mg/dl) and the results obtained were as follows :
87 71 83 67 85
77 69 76 65 85
85 54 70 68 80
73 78 68 85 73
81 78 81 77 75
Find the mean, median, and mode (mg/dl) of the above data.
72. A total of 25 patients admitted to a hospital are tested for levels of blood sugar, (mg/dl) and the results obtained were as follows :
87 71 83 67 85
77 69 76 65 85
85 54 70 68 80
73 78 68 85 73
81 78 81 77 75
Find the mean, median, and mode (mg/dl) of the above data.
Explanation:
Given, blood sugar levels are checked on a total of 25 patients admitted to a hospital.
The given data's mean, median, and mode must be determined.
all the observations are arranged in ascending order.
54 69 75 78 85
65 70 76 80 85
67 71 77 81 85
68 73 77 81 85
68 73 78 83 87
Mean is equal to the total of all observations divided by the total number of observations.
The sum of all observations is equal to 54 plus 65 plus 67 plus 68 plus 69 plus 70 plus 71 plus 73 plus 73 plus 75 + 76 + 77 + 78 + 78 + 80 + 81 + 81 + 83 + 85 + 85 + 85 + 87, which = 1891.
There have been 25 observations in total.
Mean = 1891/25 = 75.64
Given that n = 25 is odd,
[(n + 1)/2] is the median.13th observation is
= (25 + 1)/2
which is 26/2.
Therefore, the median value is 77, and the mode is the observation's maximum frequency.
The observation was repeated a maximum of 85 times; the mode was 85.
As a result, the observations' means, medians, and modes are 75.64, 77, and 85, respectively.
Chapter-14, STATISTICS AND PROBABILITY