Measures of Central Tendency — Exercise 21.1

Calculate the arithmetic mean of 5.7, 6.6, 7.2, 9.3, 6.2.

Answer

The sum of terms = 5.7 + 6.6 + 7.2 + 9.3 + 6.2 = 35

Number of terms = 5

Arithmetic mean (A.M.) = $\dfrac{\text{Sum of terms}}{\text{No. of terms}} = \dfrac{∑x_i}{n}$

$\therefore A.M. = \dfrac{35}{5} = 7.$

Hence, the mean of 5.7, 6.6, 7.2, 9.3, 6.2 is 7.

The marks obtained by 15 students in a class test are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20. Find :

(i) the mean of their marks.

(ii) the mean of their marks when the marks of each student are increased by 4.

(iii) the mean of their marks when 2 marks are deducted from the marks of each student.

(iv) the mean of their marks when the marks of each student are doubled.

Answer

(i) The sum of marks of all students = 12 + 14 + 07 + 09 + 23 + 11 + 08 + 13 + 11 + 19 + 16 + 24 + 17 + 03 + 20 = 207.

$\therefore \text{The mean of marks} = \dfrac{\text{Sum of marks of all students}}{\text{No. of students}} = \dfrac{∑x_i}{n} \\[1em] = \dfrac{207}{15} \\[1em] = 13.8$

Hence, the mean of marks of 15 students is 13.8

(ii) When the marks of each student are increased by 4, then the sum of their marks increases by 15 × 4 i.e. by 60.

∴ The new sum of marks of all students = 207 + 60 = 267.

$\therefore \text{The new mean of marks} = \dfrac{\text{ New sum of marks}}{\text{No. of students}} \\[1em] = \dfrac{267}{15} \\[1em] = 17.8$

Hence, the mean of marks of 15 students, when the marks of each student are increased by 4 is 17.8

(iii) When the marks of each student is decreased by 2, then the sum of their marks decreases by 15 × 2 i.e. by 30.

∴ The new sum of marks of all students = 207 - 30 = 177.

$\therefore \text{The new mean of marks} = \dfrac{\text{ New sum of marks}}{\text{No. of students}} \\[1em] = \dfrac{177}{15} \\[1em] = 11.8$

Hence, the mean of marks of 15 students, when the marks of each student is decreased by 2 is 11.8

(iv) When the marks of each student are doubled, then the sum of their marks will also be doubled.

∴ The new sum of marks of all students = 207 × 2 = 414.

$\therefore \text{The new mean of marks} = \dfrac{\text{ New sum of marks}}{\text{No.of students}} \\[1em] = \dfrac{414}{15} \\[1em] = 27.6$

Hence, the mean of marks of 15 students, when the marks of each student is doubled is 27.6

The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.

Answer

Arithmetic mean (A.M.) = $\dfrac{\text{Sum of terms}}{\text{No. of terms}} = \dfrac{∑x_i}{n}$

$\therefore 8 = \dfrac{6 + y + 7 + x + 14}{5} \\[1em] \Rightarrow 8 = \dfrac{x + y + 27}{5} \\[1em] \Rightarrow 40 = x + y + 27 \\[1em] \Rightarrow x + y = 13 \\[1em] \Rightarrow y = 13 - x.$

Hence, the value of y = 13 - x.

The mean of 9 variates is 11. If eight of them are 7, 12, 9, 14, 21, 3, 8 and 15, find the 9th variate.

Answer

Let 9th variate be x.

Sum of terms = 7 + 12 + 9 + 14 + 21 + 3 + 8 + 15 + x = 89 + x

Number of terms = 9.

Arithmetic mean (A.M.) = $\dfrac{\text{Sum of terms}}{\text{No. of terms}}$

Given, A.M. = 11

$\therefore 11 = \dfrac{89 + x}{9} \\[1em] \Rightarrow 99 = 89 + x \\[1em] \Rightarrow x = 99 - 89 \\[1em] \Rightarrow x = 10.$

Hence, the value of 9th variate is 10.

The mean age of 33 students of a class is 13 years. If one girl leaves the class, the mean becomes $12\dfrac{15}{16}$ years. What is the age of the girl ?

Answer

$\text{Arithmetic mean (A.M.)} = \dfrac{\text{Sum of age of students}}{\text{No.of students}} \\[1em] \Rightarrow 13 = \dfrac{\text{Sum of age of students}}{33} \\[1em] \Rightarrow \text{Sum of age of students} = 33 \times 13 \\[1em] \Rightarrow \text{Sum of age of students} = 429.$

Let the age of girl that leaves the class be x. So, sum of age of students becomes 429 - x and total no of students = 32. Given, new mean = $12\dfrac{15}{16}$.

$\therefore 12\dfrac{15}{16} = \dfrac{429 - x}{32} \\[1em] \dfrac{207}{16} = \dfrac{429 -x}{32} \\[1em] 207 \times 32 = 16(429 - x) \\[1em] 6624 = 6864 - 16x \\[1em] 16x = 6864 - 6624 \\[1em] 16x = 240 \\[1em] x = \dfrac{240}{16} = 15.$

Hence, the age of girl is 15 years.

In a class test, the mean of marks scored by a class of 40 students was calculated as 18.2. Later on, it was detected that the marks of one student was wrongly copied as 21 instead of 29. Find the correct mean.

Answer

$\text{ Mean of marks} = \dfrac{\text{Incorrect sum of marks}}{\text{No. of students}} \\[1em] \Rightarrow 18.2 = \dfrac{\text{Incorrect sum of marks}}{40} \\[1em] \Rightarrow \text{Incorrect sum of marks} = 40 \times 18.2 \\[1em] \Rightarrow \text{Incorrect sum of marks} = 728.$

As the marks of one student was wrongly copied as 21 instead of 29, correct sum of marks = 728 - 21 + 29 = 736.

$\therefore \text{Correct mean } = \dfrac{\text{Correct sum of marks}}{\text{No. of students}} \\[1em] = \dfrac{736}{40} = 18.4$

Hence, the correct mean is 18.4.

Find the mean of 25 given numbers when the mean of 10 of them is 13 and the mean of the remaining numbers is 18.

Answer

$\text{Arithmetic mean (A.M.)} = \dfrac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] \therefore \text{Sum of terms} = \text{A.M.} \times \text{No. of terms}.$

Given, mean of 10 numbers is 13.

∴ Sum of 10 terms = 13 × 10 = 130.

Given, mean of 15 numbers is 18.

∴ Sum of 15 terms = 15 × 18 = 270.

Sum of 25 terms = 130 + 270 = 400.

$\therefore A.M. = \dfrac{400}{25} = 16$

Hence, the mean of 25 numbers is 16.

Find the mean of the following distribution :

Answer

We construct the following table:

Mean = $\dfrac{∑f_ix_i}{∑f_i} = \dfrac{390}{20}$ = 19.5

Hence, the mean of the following distribution is 19.5

The contents of 100 matchboxes were checked to determine the number of matches they contained.

(i) Calculate, correct to one decimal place, the mean number of matches per box.

(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean upto exactly 39 matches.

Answer

(i) We construct the following table:

Mean = $\dfrac{∑f_ix_i}{∑f_i} = \dfrac{3813}{100}$ = 38.1

Hence, the mean of number of matches per box is 38.1

(ii) Mean = $\dfrac{\text{No. of matches}}{\text{No. of boxes}}$

Let the no. of matches added to total contents of 100 boxes be x in order to bring mean to 39. So, total matches becomes 3813 + x.

$\therefore 39 = \dfrac{3813 + x}{100} \\[1em] \Rightarrow 3900 = 3813 + x \\[1em] \Rightarrow x = 3900 - 3813 = 87.$

Hence, 87 extra matches need to added to bring the mean upto exactly 39 matches.

Find the mean for the following distribution by short cut method:

Answer

We construct the following table as under taking the assumed mean, a = 63.

Mean = $a + \dfrac{∑f_id_i}{∑f_i} = 63 + \dfrac{-23}{60} = 63- 0.38$ = 62.62.

Hence, the mean of the following distribution is 62.62.

(i) Calculate the mean wage, correct to the nearest rupee.

(ii) If the number of workers in each category is doubled, what would be the new mean wage?

Answer

(i) We construct the following table:

Mean = $\dfrac{Σf_ix_i}{Σf_i}$

= $\dfrac{42400}{50}$

= 848.

Hence, the mean wage is ₹ 848.

(ii) If the number of workers in each category is doubled then total wage will also be doubled.

New total wage = 42400 × 2 = 84800 and number of workers = 50 × 2 = 100.

Mean = $\dfrac{\text{New total wage}}{\text{No. of workers}}$

= $\dfrac{∑f_ix_i}{∑f_i}$

= $\dfrac{84800}{100}$​ = 848.

Hence, the new mean wage is also ₹ 848.

The mean of the following data is 16. Calculate the value of f.

Answer

By formula; Mean = $\dfrac{Σf_ix_i}{Σf_i}$

Substituting the values, we get

$\Rightarrow 16 = \dfrac{415 + 15f}{25 + f}\\[1em] \Rightarrow 16(25 + f) = 415 + 15f\\[1em] \Rightarrow 400 + 16f = 415 + 15f\\[1em] \Rightarrow 16f - 15f = 415 - 400\\[1em] \Rightarrow f = 15.$

Hence, the value of f = 15.

Marks obtained by 40 students in a short assessment is given below, where a and b are two missing data :

If the mean of the distribution is 7.2, find a and b.

Answer

We construct the following table:

Given, total no. of students = 40 and mean = 7.2

∴ 35 + a + b = 40
⇒ a + b = 5
⇒ a = 5 - b      .....(i)

$\text{Mean} = \dfrac{∑f_ix_i}{∑f_i} \\[1em] \therefore 7.2 = \dfrac{246 + 6a + 9b}{40} \\[1em] \Rightarrow 288 = 246 + 6a + 9b \\[1em] \Rightarrow 288 - 246 = 6a + 9b \\[1em] \Rightarrow 6a + 9b = 42$

Putting value of a from Eq (i)

$\Rightarrow 6(5 - b) + 9b = 42 \\[1em] \Rightarrow 30 - 6b + 9b = 42 \\[1em] \Rightarrow 30 + 3b = 42 \\[1em] \Rightarrow 3b = 12 \\[1em] \Rightarrow b = 4.$

a = 5 - b = 5 - 4 = 1.

Hence, the value of a = 1 and b = 4.

Calculate the mean of the following distribution:

Answer

$\text{Mean }= \dfrac{∑f_iy_i}{∑f_i}\\[1em] = \dfrac{780}{24}\\[1em] = 32.5$

Hence, mean of the following distribution is 32.5.

Calculate the mean of the following distribution :

Answer

We construct the following table:

∴ Mean = $\dfrac{∑f_iy_i}{∑f_i} = \dfrac{3600}{100}$ = 36.

Hence, mean of the following distribution is 36.

Calculate the mean of the following distribution using step deviation method :

Answer

We construct the following table, taking assumed mean a = 25.
Here, c (width of each class) = 10.

$\therefore \text{ Mean} = a + c \times \dfrac{∑f_iu_i}{∑f_i} \\[1em] = 25 + 10 \times \dfrac{63}{100} \\[1em] = 25 + \dfrac{630}{100} \\[1em] = 25 + 6.3 \\[1em] = 31.3$

Hence, mean of the following distribution is 31.3

The data on the number of patients attending a hospital in a month is given below. Find the average (mean) number of patients attending the hospital in a month using the short cut method.

Take assumed mean as 45. Give your answer correct to 2 decimal places.

Answer

Construct table as under, taking assumed mean a = 45.

$\therefore \text{ Mean} = a + \dfrac{∑f_id_i}{∑f_i} \\[1em] = 45 + \dfrac{-140}{30} \\[1em] = 45 + (-4.67) \\[1em] = 40.33$

Hence, mean of the following distribution is 40.33

The following table gives the daily wages of worker in a factory:

Calculate their mean by short cut method.

Answer

We construct the following table as under taking the assumed mean, a = 625.

By formula,

$\text{Mean }= a + \dfrac{∑f_id_i}{∑f_i}\\[1em] = 625 + \dfrac{-250}{100}\\[1em] = 625 - 2.5\\[1em] = 622.5$

Hence, mean of the following distribution is ₹622.5.

Calculate the mean of the distribution given below using the short cut method.

Answer

Construct the table as under, taking assumed mean as 45.5

$\therefore \text{ Mean} = a + \dfrac{∑f_id_i}{∑f_i} \\[1em] = 45.5 + \dfrac{70}{50} \\[1em] = 45.5 + 1.4 \\[1em] = 46.9$

Hence, mean of the following distribution is 46.9 marks

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Answer

We construct the following table :

∴ Mean = $\dfrac{∑f_iy_i}{∑f_i} = \dfrac{499}{40}$ = 12.475

Hence, mean of the following distribution is 12.475

If the mean of the following distribution is 24, find the value of a :

Answer

We construct the following table :

$\therefore \text{Mean} = \dfrac{∑f_iy_i}{∑f_i} \\[1em] \Rightarrow 24 = \dfrac{810 + 15a}{30 + a} \\[1em] \Rightarrow 24(30 + a) = 810 + 15a \\[1em] \Rightarrow 720 + 24a = 810 + 15a \\[1em] \Rightarrow 24a - 15a = 810 - 720 \\[1em] \Rightarrow 9a = 90 \\[1em] \Rightarrow a = 10.$

Hence, the value of a = 10.

The mean of the following distribution is 50. Find the unknown frequency.

Answer

By formula,

Mean = $\dfrac{∑f_ix_i}{∑f_i}$

Substituting values we get :

$\Rightarrow 50 = \dfrac{2020 + 30f}{34 + f}\\[1em] \Rightarrow 50(34 + f) = 2020 + 30f\\[1em] \Rightarrow 1700 + 50f = 2020 + 30f\\[1em] \Rightarrow 50f - 30f = 2020 - 1700\\[1em] \Rightarrow 20f = 320\\[1em] \Rightarrow f = \dfrac{320}{20}\\[1em] \Rightarrow f = 16.$

Hence, the value of f = 16.

The mean of the following frequency distribution is 57.6 and the sum of all the frequencies is 50. Find the values of p and q :

Answer

We construct the following table :

Given,

The sum of frequencies = 50.

∴ 32 + p + q = 50
⇒ p + q = 18
⇒ p = 18 - q      .....(i)

$\therefore \text{Mean} = \dfrac{∑f_iy_i}{∑f_i} \\[1em] \Rightarrow 57.6 = \dfrac{1940 + 30p + 70q}{50} \\[1em] \Rightarrow 57.6 \times 50 = 1940 + 30p + 70q \\[1em] \Rightarrow 2880 = 1940 + 30p + 70q \\[1em] \Rightarrow 30p + 70q = 2880 - 1940 \\[1em] \Rightarrow 30p + 70q = 940$

Putting value of p from (i) in above equation,

$\Rightarrow 30(18 - q) + 70q = 940 \\[1em] \Rightarrow 540 - 30q + 70q = 940 \\[1em] 540 + 40q = 940 \\[1em] 40q = 940 - 540 \\[1em] 40q = 400 \\[1em] q = 10$

Using (i),

⇒ p = 18 - q = 18 - 10 = 8.

Hence, the value of p = 8 and q = 10.

The following table gives the life time in days of 100 electricity tubes of a certain make :

Find the mean lifetime of electricity tubes.

Answer

We construct the following table :

∴ Mean = $\dfrac{∑f_iy_i}{∑f_i} = \dfrac{14500}{100}$ = 145

Hence, mean of the following distribution is 145 days

The following table gives the duration of movies in minutes.

Using step–deviation method, find the mean duration of the movies.

Answer

In the given table i is the class interval which is equal to 10.

Mean = A + $\dfrac{Σfu}{Σf} \times i = 135 + \dfrac{-36}{51} \times 10$

= $135 + \dfrac{-360}{51}$

= 135 - 7.06

= 127.94 (approx)

Hence, mean duration = 127.94 minutes (approx).

Shown below is a table illustrating the monthly income distribution in a company with 100 employees.

Using step-deviation method, find the mean monthly income of an employee.

Answer

In the given table,

Class size (i) = 4.

By formula,

Mean = A + $\dfrac{Σfu}{Σf} \times i$

= 6 + $\dfrac{19}{100} \times 4$

= 6 + $\dfrac{76}{100}$

= 6 + 0.76

= 6.76

Hence, mean = 6.76

Using the information given in the adjoining histogram, calculate the mean correct to one decimal place.

Answer

We construct the following table :

∴ Mean = $\dfrac{∑f_iy_i}{∑f_i} = \dfrac{1545}{33}$ = 46.8

Hence, mean of the following distribution is 46.8