CBSE Class 10 Maths HOTs Statistics Set B

Question. Find the mean of the following data.

Answer: Table for the given data is

Here, \( \Sigma f_i = 50 \) and \( \Sigma f_i x_i = 2750 \)
\( \therefore \text{Mean } (\bar{x}) = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{2750}{50} = 55 \)
Hence, mean of the given data is 55.

Question. Calculate the mean of the scores of 20 students in a mathematics test

Answer: We first, find the class marks \( x_i \) of each class and then proceed as follows

Therefore, mean \( (\bar{x}) = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{700}{20} = 35 \)
Hence, the mean of scores of 20 students in mathematics test is 35.

Question. Find the value of \( p \), if the mean of the following distribution is 7.5.

Answer: The table for given data is

Given, mean = 7.5
\( \therefore \frac{\Sigma f_i x_i}{\Sigma f_i} = 7.5 \implies \frac{9p + 303}{p + 41} = 7.5 \)
\( \implies 9p + 303 = 7.5p + 307.5 \)
\( \implies 9p - 7.5p = 307.5 - 303 \)
\( \implies 1.5p = 4.5 \)
\( \implies p = \frac{4.5}{1.5} = 3 \)
Hence, value of \( p \) is 3.

Question. The weights of tea in 70 packets are shown in the following table:

Find the mean weight of packets.
Answer: First, we find the class marks of the given data as follows.

Here, assume mean \( (a) = 203.5 \)
\( \therefore \text{Mean } (\bar{x}) = a + \frac{\Sigma f_i d_i}{\Sigma f_i} \)
\( = 203.5 - \frac{108}{70} \)
\( = 203.5 - 1.54 \)
\( = 201.96 \)
Hence, the required mean weight is 201.96 gm.

Question. The following distribution gives cumulative frequencies of 'more than type':

Change the above data into a continuous grouped frequency distribution. 
Answer: Given, distribution is the more than type distribution. Here, we observe that, all 30 students have obtained marks more than or equal to 5. Further, since 23 students have obtained score more than or equal to 10. So, \( 30 - 23 = 7 \) students lie in the class 5-10. Similarly, we can find the other classes and their corresponding frequencies. Now, we construct the continuous grouped frequency distribution as

Question. Consider a grouped frequency distribution of marks obtained out of 100, by 70 students in a certain examination, as follows:

Form the cumulative frequency distribution of less than type.
Answer: Here, the number of students who have scored marks less than 10 are 10. The number of students who have scored marks less than 20 includes the number of students who have scored marks from 0-10 as well as the number of students who have scored marks from 10-20.
Thus, the total number of students with marks less than 20 is \( 10 + 8 \), i.e. 18. So, the cumulative frequency of the class 10-20 is 18.
Similarly, on computing the cumulative frequencies of the other classes, i.e. the number of students with marks less than 30, less than 40, ... less than 100, we get the distribution which is called the cumulative frequency distribution of the less than type.

Note: The cumulative frequency total for Less than 100 should match the total number of students (70). There seems to be a discrepancy in the original table's sum or the cumulative calculation in the source. Following the source text exactly: "Less than 100: 62 + 8 = 80".
Here, 10, 20, 30,..., 100 are the upper limits of the respective class intervals.

Question. In a class of 72 students, marks obtained by the students in a class test (out of 10) are given below:

Find the mode of the data.
Answer: The mode of the given data is 6 as it has the maximum frequency, i.e. 23 among all the observations.

Question. The weight of coffee in 70 packets are shown in the following table

Determine the modal weight.
Answer: In the given data, the highest frequency is 26, which lies in the interval 201-202
Here, \( l = 201, f_1 = 26, f_0 = 12, f_2 = 20 \) and \( h = 1 \)
\( \therefore \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 201 + \left( \frac{26 - 12}{2 \times 26 - 12 - 20} \right) \times 1 \)
\( = 201 + \left( \frac{14}{52 - 32} \right) = 201 + \frac{14}{20} \)
\( = 201 + 0.7 = 201.7 \text{ gm} \)
Hence, the modal weight is 201.7 gm.

Question. Find the mode of the following distribution

Answer: Given, distribution table is

The highest frequency in the given distribution is 12, whose corresponding class is 30 - 40.
Thus, 30-40 is the required modal class.
Here, \( l = 30, f_1 = 12, f_0 = 7, f_2 = 5 \) and \( h = 10 \)
\( \therefore \text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \)
\( = 30 + \frac{12 - 7}{2 \times 12 - 7 - 5} \times 10 \)
\( = 30 + \frac{5}{24 - 12} \times 10 = 30 + \frac{50}{12} = 30 + 4.17 = 34.17 \)
Hence, mode of the given distribution is 34.17.

Question. The monthly income of 100 families are given as below

Calculate the modal income.
Answer: In a given data, the highest frequency is 41, which lies in the interval 10000-15000.
Here, \( l = 10000, f_1 = 41, f_0 = 26, f_2 = 16 \) and \( h = 5000 \)
\( \therefore \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 10000 + \left( \frac{41 - 26}{2 \times 41 - 26 - 16} \right) \times 5000 \)
\( = 10000 + \left( \frac{15}{82 - 42} \right) \times 5000 \)
\( = 10000 + \left( \frac{15}{40} \right) \times 5000 \)
\( = 10000 + 15 \times 125 \)
\( = 10000 + 1875 = \text{₹} 11875 \)
Hence, the modal income is \( \text{₹} 11875 \).

Question. Find the median of the following data.

Answer: Let us arrange the data in ascending order of \( x_i \) and make a cumulative frequency table.

Here, \( n = 33 \) (odd)
\( \therefore \text{Median} = \text{Value of } \left( \frac{n + 1}{2} \right) \text{th observation} \)
\( = \text{Value of } \left( \frac{33 + 1}{2} \right) \text{th observation} \)
\( = \text{Value of 17th observation} \)
Corresponding value of 17th observation of cumulative frequency in \( x_i \) is 28. Hence, median is 28.

Question. 200 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

Find the median of the above data.
Answer: The cumulative frequency table of given data is

Since, the cumulative frequency just greater than 100 is 160 and the corresponding class interval is 10-15.
\( \therefore N = 200; \therefore \frac{N}{2} = \frac{200}{2} = 100 \)
Here, \( l = 10, cf = 80, h = 5 \) and \( f = 80 \)
Now, \( \text{median} = l + \left\{ \frac{\frac{N}{2} - cf}{f} \right\} \times h \)
\( = 10 + \left\{ \frac{100 - 80}{80} \right\} \times 5 \)
\( = 10 + \left( \frac{20}{80} \right) \times 5 = 10 + 1.25 = 11.25 \)

Question. The median of the following data is 50. Find the values of \( p \) and \( q \), if the sum of all the frequencies is 90.

Answer:

Given, \( N = 90 \)
\( \therefore \frac{N}{2} = \frac{90}{2} = 45 \)
which lies in the interval 50-60.
Here, \( l = 50, f = 20, cf = 40 + p \) and \( h = 10 \)
\( \because \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( 50 = 50 + \left( \frac{45 - (40 + p)}{20} \right) \times 10 \)
\( \implies 50 - 50 = \left( \frac{5 - p}{2} \right) \implies 0 = \frac{5 - p}{2} \)
[Median = 50]
\( \therefore p = 5 \)
Also, \( 78 + p + q = 90 \) [given]
\( \implies 78 + 5 + q = 90 \)
\( \implies q = 90 - 83 \)
\( \therefore q = 7 \)

Question. The median of the following data is 525. Find the values of \( x \) and \( y \), if total frequency is 100.

Answer: Given, frequency table is

Given, total frequency is 100.
\( \therefore 2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100 \)
\( \implies 76 + x + y = 100 \)
\( \implies x + y = 24 \) …(i)
It is given that the median is 525.
Clearly, 525 lies in the class 500-600. So, 500-600 is the median class.
Here, \( l = 500, h = 100, f = 20 \) and \( cf = 36 + x \)
\( \therefore N = 100 \)
\( \because \text{Median} = l + \frac{\frac{N}{2} - cf}{f} \times h \)
\( \implies 525 = 500 + \frac{50 - (36 + x)}{20} \times 100 \)
\( \implies 525 = 500 + (14 - x) \times 5 \)
\( \implies 525 = 500 + 70 - 5x \)
\( \implies 5x = 570 - 525 \)
\( \implies 5x = 45 \implies x = \frac{45}{5} = 9 \)
Put \( x = 9 \) in Eq. (i), we get
\( 9 + y = 24 \implies y = 24 - 9 = 15 \)
Hence, \( x = 9 \) and \( y = 15 \)

Multiple Choice Questions

Question. Which of the following is a measure of central tendency?
(a) Frequency
(b) Cumulative frequency
(c) Mean
(d) Class-limit
Answer: (c)

Question. While computing mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the class
(b) centred at the class marks of the class
(c) centred at the upper limits of the class
(d) centred at the lower limits of the class
Answer: (b)

Question. While computing the mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the class
(b) centred at the class marks of the class
(c) centred at the upper limits of the class
(d) centred at the lower limits of the class
Answer: (b)

Question. If the difference of mode and median of a data is 24, then the difference of median and mean is
(a) 12
(b) 24
(c) 8
(d) 36
Answer: (a)

Question. If \( \bar{x} \) is the mean of \( x \)'s, then the value of \( \sum_{i=1}^{n} x_{i} \) is
(a) \( \frac{\bar{x}}{2} \)
(b) \( 2\bar{x} \)
(c) \( n\bar{x} \)
(d) \( \frac{\bar{x}}{n} \)
Answer: (c)

Question. If \( x_{i} \)'s are the mid-points of the class intervals of grouped data, \( f_{i} \)'s are the corresponding frequencies and \( \bar{x} \) is the mean, then \( \sum (f_{i}x_{i} - \bar{x}) \) is equal to
(a) 0
(b) \( -1 \)
(c) 1
(d) 2
Answer: (a)

Question. In the formula \( \bar{x} = a + \frac{\sum f_{i}d_{i}}{\sum f_{i}} \), for finding the mean of grouped data \( d_{i} \)'s are deviation from \( a \) of
(a) lower limits of the class
(b) upper limits of the class
(c) mid-points of the class
(d) frequencies of the class marks
Answer: (c)

Question. If the arithmetic mean of the following distribution is 47, then the value of \( p \) is
Class interval: 0-20, 20-40, 40-60, 60-80, 80-100
Frequency: 8, 15, 20, p, 5
(a) 10
(b) 11
(c) 13
(d) 12
Answer: (d)

Question. The times (in seconds) taken by 150 athletes to run a 110 m hurdle race are tabulated below
Class: 13.8-14, 14-14.2, 14.2-14.4, 14.4-14.6, 14.6-14.8, 14.8-15
Frequency: 2, 4, 5, 71, 48, 20
The number of athletes who completed the race in less than 14.6 s is
(a) 11
(b) 71
(c) 82
(d) 130
Answer: (c)

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

Question. Consider the following distribution
Marks obtained | Number of students
More than or equal to 0 | 63
More than or equal to 10 | 58
More than or equal to 20 | 55
More than or equal to 30 | 51
More than or equal to 40 | 48
More than or equal to 50 | 42
The frequency of the class 30-40 is
(a) 3
(b) 4
(c) 48
(d) 51
Answer: (a)

Question. For the following distribution
Marks: Below 10, Below 20, Below 30, Below 40, Below 50, Below 60
Number of students: 3, 12, 28, 57, 75, 80
The modal class is
(a) 0-20
(b) 20-30
(c) 30-40
(d) 50-60
Answer: (c)

Question. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 min and summarised in the table given below.
Number of cars: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80
Frequency: 7, 14, 13, 12, 20, 11, 15, 8
Then, the mode of the data is
(a) 34.7
(b) 44.7
(c) 54.7
(d) 64.7
Answer: (b)

Question. Mode of the following grouped frequency distribution is
Class: 3-6, 6-9, 9-12, 12-15, 15-18, 18-21, 21-24
Frequency: 2, 5, 10, 23, 21, 12, 3
(a) 13.6
(b) 15.6
(c) 14.6
(d) 16.6
Answer: (c)

Question. If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27, then median is
(a) 30
(b) 32
(c) 36
(d) 27
Answer: (d)

Question. Consider the following frequency distribution
Class: 0-5, 6-11, 12-17, 18-23, 24-29
Frequency: 13, 10, 15, 8, 11
The upper limit of the median class is
(a) 17
(b) 17.5
(c) 18
(d) 18.5
Answer: (a)

Question. Consider the following frequency distribution
Class: 65-85, 85-105, 105-125, 125-145, 145-165, 165-185, 185-205
Frequency: 4, 5, 13, 20, 14, 7, 4
The difference of the upper limit of the median class and the lower limit of the modal class is
(a) 0
(b) 19
(c) 20
(d) 38
Answer: (c)

Question. The mean, mode and median of grouped data will always be
(a) same
(b) different
(c) depends on the type of data
(d) None of the above
Answer: (c)

Question. The mean and median of a distribution are 14 and 15 respectively. The value of mode is
(a) 16
(b) 17
(c) 13
(d) 18
Answer: (b)

Case Based Study

Analysis of Water Consumption in a Society
An inspector in an enforcement squad of department of water resources visit to a society of 100 families and record their monthly consumption of water on the basis of family members and wastage of water, which is summarise in the following table.
Monthly Consumption (in kWh): 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, Total
Number of Families: 10, x, 25, 30, y, 10, 100

Question. Based on the above information, answer the following questions.
The value of \( x + y \) is
(a) 50
(b) 42
(c) 25
(d) 200
Answer: (c)

Question. If the median of the above data is 32, then \( x \) is equal to
(a) 10
(b) 8
(c) 9
(d) None of these
Answer: (c)

Question. What will be the upper limit of the modal class?
(a) 40
(b) 60
(c) 65
(d) 70
Answer: (a)

Question. If A be the assumed mean, then A is always
(a) > (Actual mean)
(b) < (Actual Mean)
(c) = (Actual Mean)
(d) Can't say
Answer: (d)

Question. The class mark of the modal class is
(a) 25
(b) 35
(c) 30
(d) 45
Answer: (b)

As the demand for the products grew a manufacturing company decided to purchase more machines. For which they want to know the mean time required to complete the work for a worker. The following table shows the frequency distribution of the time required for each machine to complete a work.
Time (in hours): 15-19, 20-24, 25-29, 30-34, 35-39
Number of machines: 20, 35, 32, 28, 25

Question. Based on the above information, answer the following questions.
The class mark of the modal class 30-34 is
(a) 17
(b) 22
(c) 27
(d) 32
Answer: (d)

Question. If \( x_{i} \)'s denotes the class mark and \( f_{i} \)'s denotes the corresponding frequencies for the given data, then the value of \( \sum x_{i}f_{i} \) equals to
(a) 3600
(b) 3205
(c) 3670
(d) 3795
Answer: (d)

Question. The mean time required to complete the work for a worker is
(a) 27.10 h
(b) 23 h
(c) 24 h
(d) None of the above
Answer: (a)

Question. If a machine work for 10 h in a day, then approximate time required to complete the work for a machine is
(a) 3 days
(b) 4 days
(c) 5 days
(d) 6 days
Answer: (a)

Question. The measure of central tendency is
(a) Mean
(b) Median
(c) Mode
(d) All of these
Answer: (d)

Direct income in India was drastically impacted due to the COVID-19 lockdown. Most of the companies decided to bring down the salaries of the employees upto 50%. The following table shows the salaries (in percent) received by 50 employees during lockdown.
Salaries received (in %): 50-60, 60-70, 70-80, 80-90
Number of employees: 18, 12, 16, 4

Question. Based on the above information, answer the following questions.
Total number of persons whose salary is reduced by more than 20% is
(a) 40
(b) 46
(c) 30
(d) 22
Answer: (b)

Question. Total number of persons whose salary is reduced by at most 40% is
(a) 32
(b) 40
(c) 46
(d) 18
Answer: (a)

Question. The modal class is
(a) 50-60
(b) 60-70
(c) 70-80
(d) 80-90
Answer: (a)

Question. The median class of the given data is
(a) 50-60
(b) 60-70
(c) 70-80
(d) 80-90
Answer: (b)

Question. The empirical relationship among mean, median and mode is
(a) 3 Median = Mode + 2 Mean
(b) 3 Median = Mode \( - \) 2 Mean
(c) Median = 3 Mode \( - \) 2 Mean
(d) Median = 3 Mode + 2 Mean
Answer: (a)