Measures of Central Tendency — Exercise 24(A)

The mean of 5 variables is 20. If four of them are 15, 25, 10 and 12; the fifth variable is :

1. 38

2. 60

3. 48

4. 28

Answer

Let fifth variable be x.

By formula,

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

$\therefore 20 = \dfrac{15 + 25 + 10 + 12 + x}{5} \\[1em] \Rightarrow 20 \times 5 = x + 62 \\[1em] \Rightarrow 100 = x + 62 \\[1em] \Rightarrow x = 100 - 62 = 38.$

Hence, Option 1 is the correct option.

The mean height of 5 students is 140 cm. If height of one student is 156 cm, the mean height of remaining students is :

1. 74 cm

2. 136 cm

3. 16 cm

4. 80 cm

Answer

By formula,

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

Given,

The mean height of 5 students is 140 cm.

$\therefore \text{Mean height} = \dfrac{\text{Total height}}{\text{No. of students}} \\[1em] \Rightarrow 140 = \dfrac{\text{Total height}}{5} \\[1em] \Rightarrow \text{Total height} = 140 \times 5 \\[1em] \Rightarrow \text{Total height} = 700 \text{ cm}.$

Given,

Height of one student is 156 cm, so total height of remaining students = 700 - 156 = 544 cm.

Mean height of remaining (4) students = $\dfrac{\text{Tot. ht. of 4 students}}{4} = \dfrac{544}{4}$ = 136 cm.

Hence, Option 2 is the correct option.

The mean age of eight boys is 16 years. If two more boys, with ages 18 years and 14 years, join them, the resulting mean age is :

1. 18 years

2. 14 years

3. 16 years

4. 32 years

Answer

By formula,

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

Given,

The mean age of eight boys is 16 years.

$\therefore \text{Mean age} = \dfrac{\text{Total age}}{\text{No. of boys}} \\[1em] \Rightarrow 16 = \dfrac{\text{Total age}}{8} \\[1em] \Rightarrow \text{Total age} = 16 \times 8 \\[1em] \Rightarrow \text{Total age} = 128 \text{ years}.$

Given,

Two more boys with ages 18 years and 14 years join the group.

Total age now = 128 + 18 + 14 = 160.

New mean age = $\dfrac{\text{New total age}}{\text{No. of boys}} = \dfrac{160}{10}$ = 16 years.

Hence, Option 3 is the correct option.

The mean value of 15 numbers is 20. If one of these numbers is wrongly taken as 45 instead of 15, the correct mean is :

1. 30

2. 15

3. 20

4. 18

Answer

By formula,

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

Given, the mean value of 15 numbers is 20.

$\therefore 20 = \dfrac{\text{Sum of observations}}{15} \\[1em] \Rightarrow \text{Sum of observations} = 20 \times 15 = 300.$

Given,

One of these numbers is wrongly taken as 45 instead of 15.

Correct sum of observations = 300 - 45 + 15 = 270.

$\Rightarrow \text{Correct mean} = \dfrac{\text{Correct sum of observations}}{\text{No. of observations}} \\[1em] \Rightarrow \text{Correct mean} = \dfrac{270}{15} = 18.$

Hence, Option 4 is the correct option.

The mean of the given frequency distribution is :

1. $\dfrac{30}{15} + \dfrac{20}{5}$

2. $\dfrac{30}{15} - \dfrac{20}{5}$

3. $\dfrac{550}{20}$ = 27.5

4. $\dfrac{50}{20}$ = 2.5

Answer

By formula,

Mean = $\dfrac{Σfx}{Σf} = \dfrac{550}{20}$ = 27.5

Hence, Option 3 is the correct option.

Marks obtained (in mathematics) by 9 students are given below :

60, 67, 52, 76, 50, 51, 74, 45 and 56.

(a) Find the arithmetic mean.

(b) If marks of each student be increased by 4; what will be the new value of arithmetic mean?

Answer

(a) Sum of observations = 60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56 = 531.

By formula,

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

= $\dfrac{531}{9}$ = 59.

Hence, mean = 59.

(b) We know that,

If each given number is increased or decreased by the same quantity, the value of mean is also increased or decreased by same quantity.

∴ If marks of each student is increased by 4, then mean will also increase by 4.

New mean = 59 + 4 = 63.

Hence, new mean = 63.

(a) Find the mean of 7, 11, 6, 5 and 6.

(b) If each number given in (a) is diminished by 2; find the new value of mean.

Answer

(a) Sum of observations = 7 + 11 + 6 + 5 + 6 = 35.

By formula,

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

= $\dfrac{35}{5}$ = 7.

Hence, mean = 7.

(b) We know that,

If each given number is increased or decreased by the same quantity, the value of mean is also increased or decreased by same quantity.

∴ If each number is diminished by 2, then mean will also decrease by 2.

New mean = 7 - 2 = 5.

Hence, new mean = 5.

If the mean of 6, 4, 7, a and 10 is 8, find the value of 'a'.

Answer

Sum of observations = 6 + 4 + 7 + a + 10 = a + 27.

By formula,

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

⇒ 8 = $\dfrac{a + 27}{5}$

⇒ 40 = a + 27

⇒ a = 13.

Hence, the value of a = 13.

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

Answer

Sum of observations = 6 + y + 7 + x + 14 = x + y + 27.

By formula,

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

⇒ 8 = $\dfrac{x + y + 27}{5}$

⇒ 40 = x + y + 27

⇒ x + y = 40 - 27

⇒ x + y = 13

⇒ y = 13 - x.

Hence, y = 13 - x.

The ages of 40 students are given in the following table :

Find the arithmetic mean.

Answer

We know that,

n = Σf = 40.

By formula,

Mean = $\dfrac{Σfx}{n} = \dfrac{614}{40}$ = 15.35

Hence, arithmetic mean = 15.35

If 69.5 is the mean of 72, 70, x, 62, 50, 71, 90, 64, 58 and 82 : find the value of x.

Answer

Sum of observations : 72 + 70 + x + 62 + 50 + 71 + 90 + 64 + 58 + 82 = 619 + x.

By formula,

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

⇒ 69.5 = $\dfrac{619 + x}{10}$

⇒ 695 = x + 619

⇒ x = 695 - 619

⇒ x = 76.

Hence, x = 76.

The following table gives the heights of plants in centimeter. If the mean height of plants is 60.95 cm; find the value of 'f'.

Answer

We know that,

n = Σf = 28 + f

By formula,

⇒ Mean = $\dfrac{Σfx}{n}$

⇒ 60.95 = $\dfrac{1718 + 60f}{28 + f}$

⇒ 60.95(28 + f) = 1718 + 60f

⇒ 1706.6 + 60.95f = 1718 + 60f

⇒ 60.95f - 60f = 1718 - 1706.6

⇒ 0.95f = 11.4

⇒ f = $\dfrac{11.4}{0.95}$ = 12.

Hence, f = 12.

From the data, given below, calculate the mean wage, correct to the nearest rupee.

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

(ii) If the wages per day in each category are increased by 60%; what is the new mean wage ?

(iii) If the number of workers in each category is doubled and the wages per day per worker are reduced by 40%; what would be the new mean wage ?

Answer

Mean = $\dfrac{Σfx}{Σf} = \dfrac{3360}{42}$ = 80.

(i) Original mean = $\dfrac{Σfx}{Σf}$

If no. of workers is doubled, then

New mean = $\dfrac{2Σfx}{2Σf} = \dfrac{Σfx}{Σf}$ = original mean.

∴ If the numbers of workers in each category is doubled, then new mean wage will remain same.

Hence, mean = 80.

(ii) If the wages per day in each category are increased by 60% then new mean wage also increases by 60%.

New mean = 80 + $\dfrac{60}{100} \times 80$

= 80 + 48 = 128.

Hence, new mean = 128.

(iii) There will be no change in mean due to change in number of workers.

If wages is reduced by 40% then, mean will also reduce by 40%.

New mean = 80 - $\dfrac{40}{100} \times 80$

= 80 - 32

= 48.

Hence, new mean = 48.

The contents of 100 match boxes 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 up to exactly 39 matches ?

Answer

(i)

By formula,

Mean = $\dfrac{Σfx}{Σf} = \dfrac{3813}{100}$ = 38.13 ≈ 38.1

Hence, mean = 38.1

(ii) By formula,

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

If mean = 39.

$\Rightarrow 39 = \dfrac{\text{No. of matches}}{100} \\[1em] \Rightarrow \text{No. of matches} = 3900$

No. of matches to add = 3900 - 3813 = 87.

Hence, 87 extra matches need to be added.

If the mean of the following distribution is 3, find the value of p.

Answer

By formula,

Mean = $\dfrac{Σfx}{Σf}$

Substituting values we get,

⇒ 3 = $\dfrac{6p + 87}{33}$

⇒ 99 = 6p + 87

⇒ 12 = 6p

⇒ p = $\dfrac{12}{6}$

⇒ p = 2.

Hence, p = 2.

In the following table, Σf = 200 and mean = 73. Find the missing frequencies f₁ and f₂.

Answer

Given,

⇒ Σf = 200

⇒ 86 + f₁ + f₂ = 200

⇒ f₁ + f₂ = 200 - 86

⇒ f₁ + f₂ = 114

⇒ f₁ = 114 - f₂ ........(1)

Given, mean = 73

$\Rightarrow \dfrac{Σfx}{Σf} = 73 \\[1em] \Rightarrow \dfrac{7000 + 50f_1 + 100f_2}{86 + f_1 + f_2} = 73 \\[1em] \Rightarrow \dfrac{7000 + 50f_1 + 100f_2}{86 + 114} = 73 \\[1em] \Rightarrow 7000 + 50f_1 + 100f_2 = 200 \times 73 \\[1em] \Rightarrow 7000 + 50(f_1 + 2f_2) = 14600 \\[1em] \Rightarrow 50(f_1 + 2f_2) = 14600 - 7000 \\[1em] \Rightarrow 50(f_1 + 2f_2) = 7600 \\[1em] \Rightarrow f_1 + 2f_2 = \dfrac{7600}{50} \\[1em] \Rightarrow f_1 + 2f_2 = 152 \\[1em] \Rightarrow 114 - f_2 + 2f_2 = 152 \text{ (From 1)}\\[1em] \Rightarrow 114 + f_2 = 152 \\[1em] \Rightarrow f_2 = 152 - 114 \\[1em] \Rightarrow f_2 = 38.$

Substituting value of f₂ in (1), we get :

⇒ f₁ = 114 - f₂ = 114 - 38 = 76.

Hence, f₁ = 76 and f₂ = 38.

Find the arithmetic mean (correct to nearest whole number) by using step deviation method.

Answer

Let assumed mean (A) be 25.

By formula,

Mean = A + $\dfrac{Σft}{Σf} \times i = 25 + \dfrac{-214}{384} \times 5$

= 25 - $\dfrac{1070}{384}$

= 25 - 2.786

= 22.21 ≈ 22.

Hence, mean = 22.

Find the mean (correct to one place of decimal) by using short-cut method.

Answer

Let the assumed mean (A) be 45

By formula,

Mean = A + $\dfrac{Σfd}{Σf} = 45 + \dfrac{-88}{200}$

= 45 - 0.44

= 44.56 ≈ 44.6

Hence, mean = 44.6