Mean & Median of Ungrouped Data & Frequency Polygon — Multiple Choice Questions

The mean of the numbers : 7, 9, 4, 6, 5 is :

1. 5.8

2. 6.0

3. 6.2

4. 6.4

Answer

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

= $\dfrac{7 + 9 + 4 + 6 + 5}{5}$

= $\dfrac{31}{5}$ = 6.2

Hence, option 3 is the correct option.

The median of the data 2, 7, 9, 13, 20, 22, 24, 25, 27, 28, 35, 40, is :

1. 24

2. 23

3. 25

4. 27

Answer

The data is already arranged in the ascending order.

Number of observations, n = 12, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{12}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{12}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{6th term} + \text{7th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{22} + \text{24}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{46}}{2} \\[1em]$

∴ Median = 23.

Hence, option 2 is the correct option.

Following data have been arranged in the ascending order.

29, 32, 48, 50, x, x + 2, 72, 78, 84, 95.

If the median of the data is 63, the value of x is :

1. 31

2. 62

3. 124

4. 134

Answer

Here n = 10, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{63} = \dfrac{\left(\dfrac{10}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{10}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{63} = \dfrac{\text{5th term} + \text{6th term}}{2} \\[1em] \Rightarrow \text{63} = \dfrac{\text{x} + \text{x+2}}{2} \\[1em]$

⇒ 126 = 2x + 2

⇒ 2x = 124

⇒ x = 62.

Hence, option 2 is the correct option.

If the mean of 10, 12, 18, 13, 20 and 17 is :

1. 14

2. 15

3. 16

4. 18

Answer

By arranging data in ascending order we get :

10, 12, 13, 17, 18, 20

Number of observations (n) = 6

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

= $\dfrac{10+12+13+17+18+20}{6}$

= $\dfrac{90}{6}$ = 15.

Hence, option 2 is the correct option.

If the mean of first 8 prime numbers is :

1. 9.625

2. 8.625

3. 10.625

4. 12

Answer

First 8 prime numbers are :

2, 3, 5, 7, 11, 13, 17, 19

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

= $\dfrac{2+3+5+7+11+13+17+19}{8}$

= $\dfrac{77}{8}$ = 9.625.

Hence, option 1 is the correct option.

If the mean of 10, 12, 16, 20, p and 26 is 17, then the value of p is :

1. 16

2. 18

3. 20

4. 24

Answer

Mean = 17

Number of observations = 6

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

⇒ 17 = $\dfrac{10+12+16+20+p+26}{6}$

⇒ 102 = 84 + p

⇒ p = 102 - 84

⇒ p = 18.

Hence, option 2 is the correct option.

Mean of 10 observations is 20 and that of other 15 observations is 16. Mean of all 25 observations will be :

1. 16.6

2. 18.6

3. 19.6

4. 17.6

Answer

Mean of 10 observations = 20

Mean of 15 observations = 16

Mean = $\dfrac{\text{Total sum}}{\text{Total number of observations}}$

Sum of 10 observations = 10 × 20 = 200

Sum of 15 observations = 15 × 16 = 240

Total Sum = 200 + 240 = 440

Total number of observations = 25

Mean = $\dfrac{440}{25}$ = 17.6.

Hence, option 4 is the correct option.

The height of 8 students in a class are 102 cm, 107 cm, 82 cm, 103 cm, 97 cm, 111 cm, 88 cm, and 94 cm. The mean height is :

1. 94 cm

2. 95 cm

3. 97 cm

4. 98 cm

Answer

Mean = $\dfrac{\text{Sum of all heights}}{\text{Total number of students}}$

= $\dfrac{102+107+82+103+97+111+88+94}{8}$

= $\dfrac{784}{8}$ = 98 cm.

Hence, option 4 is the correct option.

If the mean of the data x₁, x₂, x₃, ....., x_n is 'a', then the mean of the data x₁ + a, x₂ + a, ...., x_n + a is :

1. a

2. 2a

3. $\dfrac{1}{2}$a

4. $\dfrac{a}{3}$

Answer

Mean of x₁, x₂, x₃, ....., x_n is 'a'

So,

$\dfrac{x_1 + x_2 + \dots + x_n}{n} = a$

x₁ + x₂ + ..... + x_n = na

∴ $\sum x_i = na$

New observation is :

x₁ + a, x₂ + a, ...., x_n + a

$\Rightarrow \text{New Mean} = \dfrac{(x_1 + a) + (x_2 + a) + \dots + (x_n + a)}{n} \\[1em] \Rightarrow \text{New Mean} = \dfrac{(x_1 + x_2 + \dots + x_n) + (a + a + \dots + n \text{ times})}{n} \\[1em] \Rightarrow \text{New Mean} = \dfrac{\sum x_i + na}{n}$

Since, $\sum x_i = na$

∴ $\text{New Mean} = \dfrac{na + na}{n} = \dfrac{2na}{n}$ = 2a.

Hence, option 2 is the correct option.

If the mean of the data y₁, y₂, y₃, ...., y_n is 102, then the mean of the data 5y₁, 5y₂, 5y₃, ...., 5y_n is :

1. 102

2. 204

3. 510

4. 606

Answer

Mean of y₁, y₂, y₃, ...., y_n is 102

So,

$\dfrac{y_1 + y_2 + \dots + y_n}{n} = 102$

New observation is :

5y₁, 5y₂, 5y₃, ...., 5y_n

$\text{New Mean} = \dfrac{5y_1 + 5y_2 + \dots + 5y_n}{n} \\[1em] = \dfrac{5(y_1 + y_2 + \dots + y_n)}{n} \\[1em] = 5 \times \left( \dfrac{y_1 + y_2 + \dots + y_n}{n} \right)$

Since,

$\dfrac{y_1 + y_2 + \dots + y_n}{n} = 102$

∴ $\text{New Mean} = 5 \times 102$ = 510.

Hence, option 3 is the correct option.

The average score of Raman in last 5 innings is 42. If he scored 38, 41, 43 and 47 in first 4 innings, then his score in the fifth innings was :

1. 31

2. 41

3. 51

4. 61

Answer

Average score of 5 innings = 42

⇒ Total score = 42 × 5 = 210

Total score of first 4 innings = 38 + 41 + 43 + 47 = 169

Fifth innings score = 210 - 169 = 41.

Hence, option 2 is the correct option.

The mean of 25 observations was calculated to be 44. Later on it was found that two of the observations 34 and 46 were wrongly copied as 28 and 42. The correct mean will be :

1. 39.4

2. 40.4

3. 44.4

4. 43.9

Answer

Given

Wrong mean = 44

Total observations = 25

So,

Total wrong sum = 25 × 44 = 1100

Two numbers were wrongly copied:

Wrong numbers → 28 and 42

Correct numbers → 34 and 46

⇒ 34 - 28 = 6

⇒ 46 - 42 = 4

⇒ Increased sum = 6 + 4 = 10

So, the correct sum = 1100 + 10 = 1110

∴ Correct mean = $\dfrac{1110}{25}$ = 44.4

Hence, option 3 is the correct option.

The median of the following observations : 11, 12, 14, 18, (x + 4), 30, 32, 35, 41 arranged in ascending order is 24. The value of x is :

1. 19

2. 20

3. 21

4. 21.5

Answer

Given,

Median = 24

Set of numbers arranged in ascending order,

11, 12, 14, 18, (x + 4), 30, 32, 35, 41

Number of observations (n) = 9, which is odd.

By formula,

Median = $\dfrac{n + 1}{2}$ th observation

⇒ 24 = $\dfrac{9 + 1}{2}$ th observation

⇒ 24 = $\dfrac{10}{2}$th observation

⇒ 24 = 5th observation

⇒ 24 = x + 4

⇒ x = 24 - 4

⇒ x = 20.

Hence, option 2 is the correct option.

If the mean of x - 3, x - 1, 7, x, 2x - 1 and 3x - 5 is 3.5, then the median is :

1. 2.5

2. 3.5

3. 3.8

4. 3.9

Answer

Given,

Mean = 3.5

Total number of observations = 6

Mean = $\dfrac{\text{Sum of Observations}}{\text{Total number of Observations}}$

⇒ 3.5 = $\dfrac{x - 3 + x - 1 + 7 + x + 2x - 1 + 3x - 5}{6}$

⇒ 3.5 = $\dfrac{8x - 3}{6}$

⇒ 21 = 8x - 3

⇒ 8x = 24

⇒ x = 3.

∴ The observations are :

0, 2, 7, 3, 5, 4

By arranging the data in ascending order, we get :

0, 2, 3, 4, 5, 7

Number of observations (n) = 6, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{6}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{6}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{3rd term} + \text{4th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{3} + \text{4}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{7}{2} \\[1em]$

∴ Median = 3.5.

Hence, option 2 is the correct option.

The runs scored by a cricketer in last 20 innings are given below.

The mean runs of the cricketer per inning is :

1. 21

2. 24

3. 29

4. 30

Answer

Total innings = 20

Arranging the table into Runs and frequency :

0 × 1 = 0

5 × 3 = 15

17 × 6 = 102

32 × 5 = 160

61 × 3 = 183

70 × 2 = 140

⇒ Total runs = 0 + 15 + 102 + 160 + 183 + 140 = 600

⇒ Mean = $\dfrac{\text{Total Runs}}{\text{Total innings}}$

= $\dfrac{600}{20}$ = 30.

Hence, option 4 is the correct option.