CBSE Class 10 Maths HOTs Statistics Set 08

Multiple Choice Questions

Question. \( d_i \) is the deviation of \( x_i \) from assumed mean \( a \). If mean = \( x + \frac{\sum f_i d_i}{\sum f_i} \), then \( x \) is
(a) class size
(b) number of observations
(c) assumed mean
(d) None of the options
Answer: (c) assumed mean

Question. Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is
(a) 48
(b) 49
(c) 50
(d) 60
Answer: (c) 50
Sol. Sum of 100 observations = \( 100 \times 49 = 4900 \)
Correct sum = \( 4900 - [40 + 20 + 50] + [60 + 70 + 80] = 5000 \)
\( \therefore \) Correct mean = \( \frac{5000}{100} = 50 \).

Question. The A.M. of a set of 50 numbers is 38. If two numbers of the set namely 55 and 45 are discarded, the A.M. of the remaining set of numbers is _______________.
(a) 38.5
(b) 37.5
(c) 36.5
(d) 35.5
Answer: (b) 37.5
Sol. Total sum of 50 numbers = \( 50 \times 38 = 1900 \)
According to the question
Required A.M. = \( \frac{1900 - (55 + 45)}{48} = 37.5 \)

Question. Find the class marks of classes 10 – 25 and 35 – 55.
(a) 17.5, 55
(b) 18.5, 45
(c) 17.5, 45
(d) 17.5, 48
Answer: (c) 17.5, 45
Sol. Class marks of class 10 – 25 = \( \frac{10 + 25}{2} = 17.5 \)
Class marks of class 35 – 55 = \( \frac{35 + 55}{2} = 45 \)

Question. If \( \Sigma f_i = 11 \), \( \Sigma f_i x_i = 2p + 52 \) and the mean of any distribution is 6, find the value of \( p \).
(a) 5
(b) 6
(c) 7
(d) 8
Answer: (c) 7
Sol. \( \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} \)
\( \implies \) \( 6 = \frac{2p + 52}{11} \)
\( \implies \) \( 2p = 66 - 52 \)
\( \implies \) \( 2p = 14 \)
\( \implies \) \( p = 7 \)

Very Short Answer Type Questions

Question. If the mean of the following distribution is 8, find the value of \( p \):
x : 2, 4, 6, 10, p + 5
y : 3, 2, 3, 1, 2
Answer:

Mean = \( \frac{\Sigma xy}{\Sigma y} \)
\( \implies \) \( \frac{52 + 2p}{11} = 8 \)
\( \implies \) \( 2p = 88 - 52 \)
\( \implies \) \( 2p = 36 \)
\( \implies \) \( p = 18 \)

Question. In a school 85 boys and 35 girls appeared in a public examination. The mean marks of boys was found to be 40%, whereas the mean marks of girls was 60%. Determine the average marks percentage of the school.
Answer: \( \bar{x}_c = \frac{n_1 \cdot \bar{x}_1 + n_2 \cdot \bar{x}_2}{n_1 + n_2} \)
\( \implies \) \( \bar{x}_c = \frac{85 \times 40 + 35 \times 60}{85 + 35} \)
\( = \frac{3400 + 2100}{120} = \frac{5500}{120} = 45.83 \)

Question. Given that the mean of five numbers is 27. If one of the numbers is excluded, the mean gets reduced by 2. Determine the excluded number.
Answer: Mean of 5 numbers = 27
Sum of 5 numbers = \( 27 \times 5 = 135 \)
If one number is excluded, then mean of remaining 4 numbers = \( 27 - 2 = 25 \)
Sum of remaining 4 numbers = \( 25 \times 4 = 100 \)
\( \therefore \) Excluded number = \( 135 - 100 = 35 \).

Question. 20 years ago, when my parents got married, their average age was 23 years, now the average age of my family consisting of myself and my parents is 34 years. What is my present age?
Answer: Average age of parents 20 years ago = 23 years
\( \therefore \) Their total ages 20 years ago = \( 23 \times 2 = 46 \text{ years} \)
Let my age = \( x \) years
Sum of present age of parents = \( 46 + 20 \times 2 = 86 \text{ years} \)
Average age of family = \( \frac{86 + x}{3} \)
\( \implies \) \( 34 = \frac{86 + x}{3} \)
\( \implies \) \( 102 = 86 + x \)
\( \implies \) \( x = 16 \text{ years} \)

Short Answer Type Questions

Question. The mean of the following frequency distribution is 25. Find the value of \( f \).
Class : 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50
Frequency : 5, 18, 15, f, 6
Answer:

Mean = \( \frac{\Sigma f_i x_i}{\Sigma f_i} \)
\( \implies \) \( 25 = \frac{940 + 35f}{44 + f} \)
\( \implies \) \( 1100 + 25f = 940 + 35f \)
\( \implies \) \( 10f = 160 \)
\( \therefore \) \( f = \frac{160}{10} = 16 \)

Question. Find the mean of the following frequency distribution:
Class Interval : 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70
Frequency : 4, 4, 7, 10, 12, 8, 5
Total : 50
Answer: Let \( a = 35, h = 10 \)

Mean = \( a + \frac{\Sigma f_i u_i}{\Sigma f_i} \times h = 35 + \frac{16}{50} \times 10 \)
\( = 35 + 3.2 = 38.2 \)

Long Answer Type Questions

Question. The mean of the following distribution is 18. Find the frequency \( f \) of the class 19 – 21.
Answer:

\( \because \) Mean = \( \frac{\Sigma f_i x_i}{\Sigma f_i} \)
\( \implies \) \( 18 = \frac{704 + 20f}{40 + f} \)
\( \implies \) \( 18(40 + f) = 704 + 20f \)
\( \implies \) \( 720 + 18f = 20f + 704 \)
\( \implies \) \( 720 – 704 = 20f – 18f \)
\( \implies \) \( 16 = 2f \)
\( \implies \) \( f = \frac{16}{2} = 8 \)

Question. The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories of India. Find the mean percentage of female teachers by assumed mean method.
Percentage of female teachers: 15 – 25, 25 – 35, 35 – 45, 45 – 55, 55 – 65, 65 – 75, 75 – 85
Number of states/U.T.: 6, 11, 7, 4, 4, 2, 1
Answer: Here, \( a = 50 \)

\( \bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i} = 50 + \frac{-360}{35} = 39.71 \)

PRACTICE QUESTIONS

Question. A car travels from city A to city B, 120 km apart at an average speed of 50 km/h. It then makes a return trip at an average speed of 60 km/h. It covers another 120 km distance at an average speed of 40 km/h. The average speed over the entire 360 km will be
(a) \( \frac{50 + 60 + 40}{3} \) km/h
(b) \( \frac{3}{\frac{1}{50} + \frac{1}{60} + \frac{1}{40}} \) km/h
(c) \( \frac{300}{50 + 60 + 40} \) km/h
(d) None of the options
Answer: (b) \( \frac{3}{\frac{1}{50} + \frac{1}{60} + \frac{1}{40}} \) km/h

Question. Mean of \( n \) numbers \( x_1, x_2, ... x_n \) is \( m \). If \( x_n \) is replaced by \( x \), then new mean is
(a) \( m - x_n + x \)
(b) \( \frac{nm - x_n + x}{n} \)
(c) \( \frac{(n - 1)m + x}{n} \)
(d) \( \frac{m - x_n + x}{n} \)
Answer: (b) \( \frac{nm - x_n + x}{n} \)

Question. If \( \bar{x} \) is the mean of a distribution, then \( \sum f_i(x_i - \bar{x}) \) is equal to
(a) 1
(b) 2
(c) 3
(d) 0
Answer: (d) 0

Question. The mean of a set of numbers is \( \bar{x} \). If each number is multiplied by \( k \), then find the mean of the new set.
Answer: \( k\bar{x} \)

Question. If \( d_i = x_i - A \), \( \sum_{i=1}^n f_i = 25 \), \( A = 250 \), \( \bar{x} = 250 \), then find the value of \( \sum_{i=1}^n f_i d_i \).
Answer: Since \( \bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} \)
\( \implies 250 = 250 + \frac{\sum f_i d_i}{25} \)
\( \implies 0 = \frac{\sum f_i d_i}{25} \)
\( \implies \sum_{i=1}^n f_i d_i = 0 \)

Question. If the class mark of a continuous frequency distribution are 12, 14, 16, 18, ..., then find the class intervals corresponding to the class marks 16 and 22. 
Answer: The class size \( h \) is the difference between consecutive class marks: \( 14 - 12 = 2 \).
For class mark 16: Lower limit \( = 16 - \frac{2}{2} = 15 \), Upper limit \( = 16 + \frac{2}{2} = 17 \). The interval is 15 – 17.
For class mark 22: Lower limit \( = 22 - \frac{2}{2} = 21 \), Upper limit \( = 22 + \frac{2}{2} = 23 \). The interval is 21 – 23.

Question. Find the combined mean of a group of 150, if the mean of 50 students is 40 and that of other 100 students is 50.
Answer: Combined mean \( = \frac{(50 \times 40) + (100 \times 50)}{50 + 100} = \frac{2000 + 5000}{150} = \frac{7000}{150} = 46.67 \)

Question. The average weight of A, B, C is 45 kg. If the average weight of A and B be 40 kg. and that of B and C be 43 kg, find the weight of B. 
Answer: Total weight of A + B + C \( = 45 \times 3 = 135 \) kg.
Weight of A + B \( = 40 \times 2 = 80 \) kg.
Weight of B + C \( = 43 \times 2 = 86 \) kg.
(A + B) + (B + C) \( = 80 + 86 = 166 \) kg.
(A + B + C) + B \( = 166 \)
\( \implies 135 + B = 166 \)
\( \implies B = 166 - 135 = 31 \) kg.

Question. Find ‘p’ if the mean of the given data is 15.45.
Class: 0 – 6, 6 – 12, 12 – 18, 18 – 24, 24 – 30
Frequency: 6, 8, p, 9, 7
Answer: Class marks (\( x_i \)): 3, 9, 15, 21, 27.
\( \sum f_i = 6 + 8 + p + 9 + 7 = 30 + p \)
\( \sum f_i x_i = (6 \times 3) + (8 \times 9) + (p \times 15) + (9 \times 21) + (7 \times 27) = 18 + 72 + 15p + 189 + 189 = 468 + 15p \)
Mean \( = \frac{\sum f_i x_i}{\sum f_i} \)
\( \implies 15.45 = \frac{468 + 15p}{30 + p} \)
\( \implies 15.45(30 + p) = 468 + 15p \)
\( \implies 463.5 + 15.45p = 468 + 15p \)
\( \implies 0.45p = 4.5 \)
\( \implies p = 10 \)

Question. Find the mean of the following data using assumed mean method: 
Class: 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25
Frequency: 8, 7, 10, 13, 12
Answer: Let Assumed Mean \( a = 12.5 \)
Class | \( f_i \) | \( x_i \) | \( d_i = x_i - 12.5 \) | \( f_i d_i \)
0 – 5 | 8 | 2.5 | -10 | -80
5 – 10 | 7 | 7.5 | -5 | -35
10 – 15 | 10 | 12.5 | 0 | 0
15 – 20 | 13 | 17.5 | 5 | 65
20 – 25 | 12 | 22.5 | 10 | 120
Total \( \sum f_i = 50 \), \( \sum f_i d_i = 70 \)
Mean \( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 12.5 + \frac{70}{50} = 12.5 + 1.4 = 13.9 \)

Question. Compute mean of the grouped data:
Daily wages: 12.5 – 17.5, 17.5 – 22.5, 22.5 – 27.5, 27.5 – 32.5, 32.5 – 37.5, 37.5 – 42.5, 42.5 – 47.5, 47.5 – 52.5, 52.5 – 57.5
Number of workers: 2, 22, 19, 14, 3, 4, 6, 1, 1
Answer: Class Marks (\( x_i \)): 15, 20, 25, 30, 35, 40, 45, 50, 55
Total workers (\( \sum f_i \)) = 72
\( \sum f_i x_i = (2 \times 15) + (22 \times 20) + (19 \times 25) + (14 \times 30) + (3 \times 35) + (4 \times 40) + (6 \times 45) + (1 \times 50) + (1 \times 55) \)
\( \sum f_i x_i = 30 + 440 + 475 + 420 + 105 + 160 + 270 + 50 + 55 = 2005 \)
Mean \( = \frac{2005}{72} = 27.85 \)

Multiple-Choice Questions 

Question. For the following distribution:
Marks | Number of Students
Below 10 | 3
Below 20 | 12
Below 30 | 27
Below 40 | 57
Below 50 | 75
Below 60 | 80
the modal class is
(a) 10 – 20
(b) 20 – 30
(c) 30 – 40
(d) 50 – 60
Answer: (c) 30 – 40
Sol. (c) The modal class is 30–40.

Question. A set of numbers consists of four 5’s, six 7’s, ten 9’s, eleven 12’s, three 13’s, two 14’s. The mode of this set of numbers is
(a) 7
(b) 9
(c) 12
(d) 13
Answer: (c) 12
Sol. (c) Frequency of 12 is more than others. So, the mode of this set is 12.

Question. The mode of the numbers 2, 3, 4, 4, 3, 5, 3, 6 is
(a) 3
(b) 3.5
(c) 4
(d) 4.5
Answer: (a) 3
Sol. (a) Here, 3 comes more than others. Therefore mode = 3.

Question. Find the value of \( x \), if the mode of the following data is 25.
15, 20, 25, 18, 14, 15, 25, 15, 18, 16, 20, 25, 20, \( x \), 18
(a) 18
(b) 20
(c) 25
(d) None of the options
Answer: (c) 25
Sol. (c) Observations | Tally Marks | \( f \)
14 | | | 1
15 | ||| | 3
16 | | | 1
18 | ||| | 3
20 | ||| | 3
25 | ||| | 3
Observations 15, 18, 20, and 25 all have the same frequency i.e., 3. For 25 be the mode of the data it should have the maximum frequency.
\( \therefore \) 25 should repeat itself at least once more i.e., \( x \) should be 25.

Very Short Answer Type Question 

Question. (i) Find the value of \( x \), if the mode of the following data is 25:
15, 20, 25, 18, 14, 15, 25, 15, 18, 16, 20, 25, 20, \( x \), 18
(ii) If both 20 and 18 are changed to 15 in the above data, find the new mode.
Answer: Sol. (i) Mode of given data is 25.
\( \therefore \) 25 has maximum frequency. Hence, the value of \( x \) is 25.
(ii) If 20 and 18 both are changed to 15. Then frequency of 15 becomes the maximum i.e., 9 and new mode is 15.

Short Answer Type Question 

Question. Find the mode of the following distribution:
Class Interval | Frequency
0 – 10 | 5
10 – 20 | 8
20 – 30 | 7
30 – 40 | 12
40 – 50 | 28
50 – 60 | 20
60 – 70 | 10
70 – 80 | 10
Answer: Sol. Modal Class 40 – 50, Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( l = \) lower limit of the modal class;
\( f_1 = \) frequency of the modal class
\( f_0 = \) frequency of the class preceding the modal class;
\( f_2 = \) frequency of the class succeeding the modal class.
\( h = \) Class size
Mode \( = 40 + \left( \frac{28 - 12}{2 \times 28 - 12 - 20} \right) \times 10 \)
\( = 40 + \left( \frac{16}{56 - 32} \right) \times 10 \)
\( = 40 + \left( \frac{16}{24} \right) \times 10 \)
\( = 40 + \frac{20}{3} = 46.67 \)

Long Answer Type Question 

Question. The ages of employees in two factories A and B are given below:
Age of Employees (in years) | Number of employees in factories A | Number of employees in factories B
20 – 30 | 5 | 8
30 – 40 | 26 | 40
40 – 50 | 78 | 58
50 – 60 | 104 | 90
60 – 70 | 98 | 83
Find the modal age of employees in factory A and factory B.
Answer: Sol.
Age of Employees (in years) C.I. | Number of employees in factories A | Number of employees in factories B
20 – 30 | 5 | 8
30 – 40 | 26 | 40
40 – 50 | 78 | 58
50 – 60 | 104 | 90
60 – 70 | 98 | 83
For factory A, max. frequency \( = 104 \).
\( \therefore \) Modal class \( = 50 – 60 \).
Mode \( = l + \left[ \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right] \times h \)
\( f_1 = \) frequency of modal class \( = 100 \)
\( f_0 = \) frequency of preeceding class \( = 78 \)
\( f_2 = \) frequency of succeding class \( = 98 \)
\( h = \) class size \( = 10 \)
\( l = \) lower limit of the modal class \( = 50 \)
Mode \( = 50 + \left[ \frac{104 - 78}{2 \times 104 - 78 - 98} \right] \times 10 \)
\( = 50 + \frac{26}{32} \times 10 = 50 + 8.125 = 58.125 \)
For factory B, max. frequency \( = 90 \).
\( \therefore \) Modal class \( = 50 – 60 \).
Mode \( = 50 + \left( \frac{90 - 58}{2 \times 90 - 58 - 83} \right) \times 10 \)
\( = 50 + \frac{32 \times 10}{39} = 50 + 8.205 \)
\( = 58.205 \)

Question. Mode is the value of the variable which has:
(a) maximum frequency
(b) minimum frequency
(c) mean frequency
(d) middle most frequency
Answer: (a) maximum frequency

Question. The shirt sizes worn by a group of 200 persons, who bought the shirt from a store are as follows:
Shirt size | No. of persons
37 | 15
38 | 25
39 | 39
40 | 41
41 | 36
42 | 17
43 | 15
44 | 12
Find the modal shirt size worn by the group.
Answer: The modal shirt size is the one with the highest frequency. In the given data, the frequency 41 is the highest, which corresponds to shirt size 40. Therefore, the modal shirt size is 40.

Question. Find the modal class for the following distribution
Marks | Number of students
Below 10 | 3
Below 20 | 12
Below 30 | 27
Below 40 | 57
Below 50 | 75
Below 60 | 80
Answer: To find the modal class, we first determine the individual frequencies for each class interval:
0 – 10: 3
10 – 20: \( 12 - 3 = 9 \)
20 – 30: \( 27 - 12 = 15 \)
30 – 40: \( 57 - 27 = 30 \)
40 – 50: \( 75 - 57 = 18 \)
50 – 60: \( 80 - 75 = 5 \)
The maximum frequency is 30, which corresponds to the class interval 30 – 40. Thus, the modal class is 30 – 40.

Question. Find the mode of the following frequency distribution 
Class | Frequency
0 – 6 | 7
6 – 12 | 5
12 – 18 | 10
18 – 24 | 12
24 – 30 | 6
Answer: Here, the modal class is 18 – 24 as it has the maximum frequency (\( f_1 = 12 \)).
\( l = 18, f_0 = 10, f_2 = 6, h = 6 \)
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 18 + \left( \frac{12 - 10}{2(12) - 10 - 6} \right) \times 6 \)
\( = 18 + \left( \frac{2}{24 - 16} \right) \times 6 \)
\( = 18 + \frac{2}{8} \times 6 \)
\( = 18 + 1.5 = 19.5 \)

Question. Compare the modal ages of two groups of students A and B appearing for an entrance test:
Age (in years) | Number of students in Group A | Number of students in Group B
16 – 18 | 50 | 54
18 – 20 | 78 | 89
20 – 22 | 46 | 40
22 – 24 | 28 | 25
24 – 26 | 23 | 17
Answer: For Group A:
Modal class = 18 – 20 (\( f_1 = 78, f_0 = 50, f_2 = 46, l = 18, h = 2 \))
Mode \( = 18 + \left( \frac{78 - 50}{2(78) - 50 - 46} \right) \times 2 = 18 + \frac{28}{156 - 96} \times 2 = 18 + \frac{56}{60} \approx 18.93 \)
For Group B:
Modal class = 18 – 20 (\( f_1 = 89, f_0 = 54, f_2 = 40, l = 18, h = 2 \))
Mode \( = 18 + \left( \frac{89 - 54}{2(89) - 54 - 40} \right) \times 2 = 18 + \frac{35}{178 - 94} \times 2 = 18 + \frac{70}{84} \approx 18.83 \)
The modal age for Group A (\( 18.93 \)) is slightly higher than for Group B (\( 18.83 \)).

Question. If mode of the following frequency distribution is 55, then find the value of x.
Class | Frequency
0 – 15 | 10
15 – 30 | 7
30 – 45 | x
45 – 60 | 15
60 – 75 | 10
75 – 90 | 12
Answer: Given Mode = 55. This lies in the class 45 – 60, so modal class is 45 – 60.
\( l = 45, f_1 = 15, f_0 = x, f_2 = 10, h = 15 \)
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( \implies 55 = 45 + \left( \frac{15 - x}{2(15) - x - 10} \right) \times 15 \)
\( \implies 10 = \left( \frac{15 - x}{20 - x} \right) \times 15 \)
\( \implies \frac{10}{15} = \frac{15 - x}{20 - x} \)
\( \implies \frac{2}{3} = \frac{15 - x}{20 - x} \)
\( \implies 2(20 - x) = 3(15 - x) \)
\( \implies 40 - 2x = 45 - 3x \)
\( \implies x = 5 \)

Question. Find the unknown entries a, b, c, d, e and f in the following distribution and hence find their mode. 
Height (in cm) | Frequency | Cumulative Frequency
150 – 155 | 12 | a
155 – 160 | b | 25
160 – 165 | 10 | c
165 – 170 | d | 43
170 – 175 | e | 48
175 – 180 | 2 | f
Total | 50 |
Answer: Comparing frequencies with cumulative frequencies:
\( a = 12 \)
\( a + b = 25 \implies 12 + b = 25 \implies b = 13 \)
\( 25 + 10 = c \implies c = 35 \)
\( c + d = 43 \implies 35 + d = 43 \implies d = 8 \)
\( 43 + e = 48 \implies e = 5 \)
\( 48 + 2 = f \implies f = 50 \)
Highest frequency is \( b = 13 \), so modal class is 155 – 160.
\( l = 155, f_1 = 13, f_0 = 12, f_2 = 10, h = 5 \)
Mode \( = 155 + \left( \frac{13 - 12}{2(13) - 12 - 10} \right) \times 5 = 155 + \frac{1}{4} \times 5 = 155 + 1.25 = 156.25 \)