Measures of Central Tendency (Mean) — Exercise 25

Find the mean of each of the following sets of numbers :

(i) 10, 4, 6, 12, 9

(ii) 0.2, 0.02, 2, 2.02, 1.22, 1.02

Answer

(i) Given,

10, 4, 6, 12, 9

We know that,

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{10 + 4 + 6 + 12 + 9}{5} \\[1em] = \dfrac{41}{5} \\[1em] = 8.2$

Hence, mean of given numbers = 8.2.

(ii) Given,

0.2, 0.02, 2, 2.02, 1.22, 1.02

We know that,

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{0.2 + 0.02 + 2 + 2.02 + 1.22 + 1.02}{6} \\[1em] = \dfrac{6.48}{6} \\[1em] = 1.08.$

Hence, mean of given numbers = 1.08.

Find the arithmetic mean of :

(i) first eight natural numbers;

(ii) first five prime numbers;

(iii) first six positive even integers;

(iv) first five positive integral multiples of 3;

(v) all factors of 20.

Answer

(i) Given,

first eight natural numbers = 1, 2, 3, 4, 5, 6, 7, 8

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}{8} \\[1em] = \dfrac{36}{8} \\[1em] = 4.5$

Hence, mean of given numbers = 4.5.

(ii) Given,

first five prime numbers = 2, 3, 5, 7, 11

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{2 + 3 + 5 + 7 + 11}{5} \\[1em] = \dfrac{28}{5} \\[1em] = 5.6$

Hence, mean of given numbers = 5.6.

(iii) Given,

first six positive even integers; = 2, 4, 6, 8, 10, 12

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{2 + 4 + 6 + 8 + 10 + 12}{6} \\[1em] = \dfrac{42}{6} \\[1em] = 7$

Hence, mean of given numbers = 7.

(iv) Given,

first five positive integral multiples of 3 = 3, 6, 9, 12, 15

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{3 + 6 + 9 + 12 + 15}{5} \\[1em] = \dfrac{45}{5} \\[1em] = 9.$

Hence, mean of given numbers = 9.

(v) Given,

all factors of 20 = 1, 2, 4, 5, 10, 20

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{1 + 2 + 4 + 5 + 10 + 20}{6} \\[1em] = \dfrac{42}{6} \\[1em] = 7.$

Hence, mean of given numbers = 7.

The daily minimum temperature recorded (in degrees F) at a place during a week was as under :

Find the mean temperature of the week.

Answer

We know that,

Mean = $\dfrac{\sum x_i}{n}$

We have,

$\Rightarrow \text{Mean of temperatures } = \dfrac{\text{Sum of temperatures of all days}}{\text{Number of days}} \\[1em] = \dfrac{35.5 + 30.8 + 28.3 + 31.1 + 23.8 + 29.9 + 32.7}{7} \\[1em] = \dfrac{212.1}{7} \\[1em] = 30.3 ^{\circ} F$

Hence, mean temperature of the week is 30.3 °F.

The marks obtained by 10 students in a class-test were as follows :

38, 41, 36, 31, 45, 38, 27, 32, 29, 39

Find :

(i) the mean of their marks;

(ii) the mean of their marks, when the marks of each student are increased by 2;

(iii) the mean of their marks, when 1 mark is deducted from the marks of each student;

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

Answer

(i) We know that,

Mean = $\dfrac{\sum x_i}{n}$

We have,

$\Rightarrow \text{Mean of marks} = \dfrac{\text{Sum of marks}}{\text{Number of students}} \\[1em] = \dfrac{38 + 41 + 36 + 31 + 45 + 38 + 27 + 32 + 29 + 39}{10} \\[1em] = \dfrac{356}{10} \\[1em] = 35.6$

Hence, mean of marks = 35.6.

(ii) If 2 marks are added to each student, the mean also increases by 2.

New mean = 35.6 + 2 = 37.6

Hence, mean when marks are increased by 2 = 37.6

(iii) If 1 mark is deducted to each student, the mean also decreases by 1.

New mean = 35.6 - 1 = 34.6

Hence, mean when 1 mark is deducted = 34.6

(iv) If the marks of each student are halved, the mean is also halved.

New mean = $\dfrac{35.6}{2}$ = 17.8

Hence, mean when marks are halved = 17.8

If the mean of 11, 8, 13, 10, x and 9 is 9.5, find the value of x.

Answer

We know that,

Mean = $\dfrac{\sum x_i}{n}$

We have,

$\Rightarrow \text{Mean} = \dfrac{11 + 8 + 13 + 10 + x + 9}{6} \\[1em] \Rightarrow 9.5 = \dfrac{51 + x}{6} \\[1em] \Rightarrow 6(9.5) = 51 + x \\[1em] \Rightarrow 57 = 51 + x \\[1em] \Rightarrow x = 57 - 51 \\[1em] \Rightarrow x = 6.$

Hence, value of x is 6.

Find the mean of 25 numbers, it being given that the mean of 15 of them is 18 and the mean of remaining ones is 13.

Answer

Mean = $\dfrac{\text{Sum of terms}}{\text{Number of terms}}$

∴ Sum of terms = Mean × Number of terms

Given, mean of 15 numbers is 18

∴ Sum of 15 terms = 18 × 15 = 270

Given, mean of 10 numbers is 13

∴ Sum of 13 terms = 13 × 10 = 130

Sum of 25 terms = 130 + 270 = 400.

Mean = $\dfrac{400}{25} = 16$

Hence, the mean of 25 numbers is 16.

The mean weight of 60 students of a class is 52.75 kg. If the mean weight of 25 of them is 51 kg, find the mean weight of the remaining students.

Answer

By formula,

Mean = $\dfrac{\text{Sum of weight of students}}{\text{No.of students}}$

Given,

Mean weight of 60 students of a class = 52.75 kg

$\therefore 52.75 = \dfrac{\text{Sum of weight of students}}{60}$

Total weight = 60 × 52.75

= 3165 kg.

Mean weight of 25 students among them = 51 kg.

So, the total weight of 25 students = 51 × 25 = 1275 kg.

Remaining students = 60 – 25 = 35

Total weight of remaining 35 students = 3165 – 1275 = 1890 kg

Mean weight of 35 students = $\dfrac{1890}{35}$ = 54 kg.

Hence, the mean weight of the remaining students is 54 kg.

The mean of five numbers is 18. On excluding one number, the mean becomes 16. Find the excluded number.

Answer

Given:

Number of observations = 5

Mean = 18

⇒ Sum of all 5 observations = 5 x 18 = 90

On excluding an observation, the mean of the remaining 4 observations = 16

∵ Sum of all remaining 4 observations = 4 x 16 = 64

⇒ Excluded observation = Sum of all 5 observations - Sum of all remaining 4 observations

= 90 - 64

= 26

Hence, the excluded number is 26.

The ages of 40 students of a group are given below :

Find the mean age of the group.

Answer

We know that,

n = ∑f = 40.

By formula,

Mean = $\dfrac{\sum fx}{n} = \dfrac{580}{40}$ = 14.5 years

Hence, mean age of the group is 14.5 years.

Find the mean of the following frequency distribution :

Answer

We know that,

n = ∑f = 50.

By formula,

Mean = $\dfrac{\sum fx}{n} = \dfrac{359}{50}$ = 7.18

Hence, mean of the frequency distribution is 7.18.

In a book of 300 pages, the distribution of misprints is shown below :

Find the average number of misprints per page.

Answer

We know that,

n = ∑f = 300.

By formula,

Mean = $\dfrac{\sum fx}{n} = \dfrac{222}{300}$ = 0.74 per page.

Hence, average number of misprints per page is 0.74.

The following table gives the wages of different categories of workers in a factory :

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

Answer

(i) We know that,

n = ∑f = 50.

By formula,

Mean wage = $\dfrac{\sum fx}{n} = \dfrac{21200}{50}$ = ₹ 424

Hence, mean wage is ₹ 424.

(ii) If all frequencies in a distribution are multiplied by a constant, the mean remains unchanged.

n = ∑f = 50 x 2 = 100.

∑fx = 21200 x 2 = 42400

By formula,

Mean wage = $\dfrac{\sum fx}{n} = \dfrac{42400}{100}$ = ₹ 424

Hence, new mean wage is ₹ 424.

If the mean of the following distribution is 7.5, find the missing frequency f :

Answer

We know that,

n = ∑f = 74 + f.

By formula,

$\text{Mean} = \dfrac{\sum fx}{n} \\[1em] 7.5 = \dfrac{563 + 7f}{74 + f}.$

⇒ 7.5(74 + f) = 563 + 7f

⇒ 555 + 7.5f = 563 + 7f

⇒ 7.5f - 7f = 563 - 555

⇒ 0.5f = 8

⇒ f = $\dfrac{8}{0.5}$

⇒ f = 16.

Hence, missing frequency f is 16.

If the mean of the following observations is 16.6, find the numerical value of p.

Answer

We know that,

n = ∑f = 76 + p.

By formula,

$\text{Mean} = \dfrac{\sum fx}{n} \\[1em] 16.6 = \dfrac{1228 + 18p}{76 + p}.$

⇒ 16.6(76 + p) = 1228 + 18p

⇒ 1261.6 + 16.6p = 1228 + 18p

⇒ 1261.6 - 1228 = 18p - 16.6p

⇒ 33.6 = 1.4p

⇒ p = $\dfrac{33.6}{1.4}$

⇒ p = 24.

Hence, numerical value of p is 24.

Find the numerical value of x, if the mean of the following frequency distribution is 12.58.

Answer

We know that,

n = ∑f = 50.

By formula,

$\text{Mean} = \dfrac{\sum fx}{n} \\[1em] 12.58 = \dfrac{524 + 7x}{50}.$

⇒ 12.58(50) = 524 + 7x

⇒ 629 = 524 + 7x

⇒ 7x = 629 - 524

⇒ 7x = 105

⇒ x = $\dfrac{105}{7}$

⇒ x = 15.

Hence, numerical value of x is 15.

Using short cut method, compute the mean height from the following frequency distribution :

Answer

Let assumed mean (A) = 62.

We know that,

n = ∑f = 80.

By formula,

$\text{Mean} = A + \dfrac{\sum fd}{n} \\[1em] = 62 + \dfrac{28}{80} \\[1em] = 62 + 0.35 \\[1em] = 62.35.$

Hence, mean height of the plants is 62.35 cm.

The number of match sticks contained in 50 match boxes is given below :

(i) Using short cut method, find the mean number of match sticks per box.

(ii) How many extra match sticks are to be added to all the contents of 50 match boxes to bring the mean exactly equal to 45 match sticks per box?

Answer

(i) We know that,

n = ∑f = 50.

By formula,

$\text{Mean} = A + \dfrac{\sum fd}{n} \\[1em] = 44 + \dfrac{-16}{50} \\[1em] = 44 - 0.32 \\[1em] = 43.68$

Hence, mean number of match sticks per box is 43.68

(ii) Total number of sticks = Mean × Total Boxes

= 43.68 × 50

= 2184 sticks

Number of match sticks to be added = (50 × 45) − 2184 = 66.

Hence, extra match sticks to be added = 66.

The following table gives the marks scored by a set of students in an examination. Calculate the mean of the distribution by using the short cut method.

Answer

By formula,

$\text{Mean} = A + \dfrac{\sum fd}{\sum f} \\[1em] = 25 + \dfrac{90}{40} \\[1em] = 25 + 2.25 \\[1em] = 27.25.$

Hence, required mean = 27.25

Using short-cut method, find mean of the given frequency distribution:

Answer

Construct the table as under, taking assumed mean as 45:

By formula,

$\text{Mean} = a + \dfrac{\sum f_id_i}{\sum f_i}$

$= 45 + \dfrac{150}{50}$

= 45 + 3

= 48.

Hence, required mean = 48.

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

Answer

By formula,

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

Hence, the value of a = 10.

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.

By formula,

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

Hence, mean of the following distribution is 31.3

Using step-deviation method, find mean for the following frequency distribution

Answer

In the given table i is the class interval which is equal to 15.

Mean = A + $\dfrac{Σfu}{Σf} \times i = 52.5 + \dfrac{-12}{30} \times 15$

= $52.5 - \dfrac{180}{30}$

= 52.5 - 6

= 46.50

Hence, mean = 46.50

The weights of 50 apples were recorded as given below. Calculate the mean weight, to the nearest gram, by the step Deviation Method.

Answer

We construct the following table, taking assumed mean a = 97.5. Here, c (width of each class) = 5.

By formula,

$\text{Mean} = a + c \times \dfrac{\sum f_iu_i}{\sum f_i} \\[1em] = 97.5 + 5 \times \dfrac{-16}{50} \\[1em] = 97.5 - \dfrac{16}{10} \\[1em] = 97.5 - 1.6 \\[1em] = 95.9 \approx 96$

Hence, mean weight of the apples is 96 g.

Weights of 60 eggs were recorded as given below :

Calculate their mean weight to the nearest gm.

Answer

Since, class are discontinuous we will first convert them into continuous class intervals.

Adjustment factor

= $\dfrac{\text{Lower limit of a class -Upper limit of previous class}}{2} = \dfrac{80 - 79}{2} = \dfrac{1}{2}$ = 0.5

Adding the adjustment factor to upper limit and subtracting from lower limit we get the continuous class intervals.

We construct the following table, taking assumed mean a = 92. Here, c (width of each class) = 5.

By formula,

$\text{Mean} = a + c \times \dfrac{\sum f_iu_i}{\sum f_i} \\[1em] = 92 + 5 \times \dfrac{-19}{60} \\[1em] = 92 - \dfrac{19}{12} \\[1em] = 92 - 1.583 \\[1em] = 90.416 \approx 90$

Hence, mean weight of the eggs is 90 g.

The following table gives marks scored by students in an examination :

Calculate the mean marks correct to 2 decimal places.

Answer

We construct the following table, taking assumed mean a = 17.5. Here, c (width of each class) = 5.

By formula,

$\text{Mean} = a + c \times \dfrac{\sum f_iu_i}{\sum f_i} \\[1em] = 17.5 + 5 \times \dfrac{17}{80} \\[1em] = 17.5 + \dfrac{85}{80} \\[1em] = 17.5 + 1.0625 \\[1em] = 18.5625 \approx 18.56$

Hence, mean marks scored by the students is 18.56

The data on the number of patients attending a hospital in a month are given below. Find the average (mean) number of patients attending the hospital in a month by using the shortcut method.

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

Answer

We construct the following table, taking assumed mean a = 45.

By formula,

$\text{Mean} = a + \dfrac{\sum f_id_i}{\sum f_i} \\[1em] = 45 + \dfrac{-140}{30} \\[1em] = 45 - \dfrac{14}{3} \\[1em] = 45 - 4.6666 \\[1em] = 40.3333 \approx 40.33$

Hence, average number of patients attending the hospital per month is 40.33.

Calculate the mean of the following frequency distribution.

Answer

We construct the following table, taking assumed mean a = 30. Here, c (width of each class) = 10.

By formula,

$\text{Mean} = a + c \times \dfrac{\sum f_iu_i}{\sum f_i} \\[1em] = 30 + 10 \times \dfrac{6}{24} \\[1em] = 30 + \dfrac{10}{4} \\[1em] = 30 + 2.5 \\[1em] = 32.5$

Hence, mean of the frequency distribution is 32.5.