Mean & Median — Test Yourself

If each observation of a given set of data is increased by 5, their mean:

1. remains the same

2. becomes five times

3. is decreased by 5

4. is increased by 5

Answer

Let x₁, x₂, x₃ be the observations.

So, their mean (M) = $\dfrac{x_1 + x_2 + x_3}{3}$

Since, each observations is increased by 5, so observations will be x₁ + 5, x₂ + 5, x₃ + 5.

$\text{New median}= \dfrac{x_1 + 5 + x_2 + 5 + x_3 + 5}{3}\\[1em] = \dfrac{x_1 + x_2 + x_3 + 15}{3}\\[1em] = \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{15}{3}\\[1em] = \dfrac{x_1 + x_2 + x_3}{3} + 5\\[1em] = M + 5.$

Hence, option 4 is the correct option.

The median of observations (written in ascending order) 26, 29, 42, 53, x, x + 2, 70, 75, 82, 83 and 100 is 65; then:

1. x = 65

2. x + 2 = 65

3. $\dfrac{x + (x + 2)}{2}$ = 6

4. none of these

Answer

Number of observations, n = 11 (odd)

By formula,

Median = $\Big[\dfrac{n + 1}{2}\Big]^{th}$ term

$\Rightarrow 65 = \Big[\dfrac{11 + 1}{2}\Big]^{th} \text{ term}\\[1em] \Rightarrow 65 = \Big[\dfrac{12}{2}\Big]^{th} \text{ term}\\[1em] \Rightarrow 65 = 6^{th} \text{ term}\\[1em] \Rightarrow 65 = x + 2.$

Hence, option 2 is the correct option.

The mean of x - 5, x - 3, x - 1 and x + 1 is :

1. 4x - 8

2. $\dfrac{4x - 8}{4}$

3. $\dfrac{x - 3 + x - 1}{2}$

4. none of these

Answer

Given, observations = x - 5, x - 3, x - 1 and x + 1

Number of observations = 4

By formula,

Mean (M) = $\dfrac{\text{Sum of observations}}{\text{Number of observation}}$

$M = \dfrac{x - 5 + x - 3 + x - 1 + x + 1}{4}\\[1em] = \dfrac{4x - 8}{4}.$

Hence, option 2 is the correct option.

26, x + 4, x + 2 and 18 are in descending order of their values and their median is 20, then the value of x is :

1. 19

2. 17

3. 15

4. 13

Answer

Given observations: 26, x + 4, x + 2 and 18

Median = 20

Arranging the above observations in ascending order,

18, x + 2, x + 4, 26

Number of observations, n = 4 (even)

By formula,

Median = $\dfrac{1}{2}\Big[\text{the value of }\Big(\dfrac{n}{2}\Big)^{th} \text{ term } + \text{the value of }\Big(\dfrac{n}{2} + 1\Big)^{th} \text{ term }\Big]$

Substituting values we get :

$\Rightarrow 20 = \dfrac{1}{2}\Big[\text{the value of }\Big(\dfrac{4}{2}\Big)^{th} \text{ term } + \text{the value of }\Big(\dfrac{4}{2} + 1\Big)^{th} \text{ term }\Big] \\[1em] \Rightarrow 20 = \dfrac{1}{2}\Big[\text{2nd term + 3rd term}\Big]\\[1em] \Rightarrow 20 = \dfrac{1}{2}\Big[x + 2 + x + 4\Big]\\[1em] \Rightarrow 20 = \dfrac{1}{2}\Big[2x + 6 \Big]\\[1em] \Rightarrow 20 = x + 3\\[1em] \Rightarrow x = 20 - 3\\[1em] \Rightarrow x = 17.$

Hence, option 2 is the correct option.

Statement 1: For n number of data in a set, median = $\Big(\dfrac{n}{2}\Big)^{th}$ term.

Statement 2: If n is even, median = $\dfrac{1}{2}\Big[\Big(\dfrac{n}{2}\Big)^{th} \text{ term } + \Big(\dfrac{n}{2}+ 1\Big)^{th} \text{ term }\Big]$

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

If number of observations (n) is odd.

Median = $\Big[\dfrac{n + 1}{2}\Big]^{th}$ term

If number of observations (n) is even.

Median = $\dfrac{1}{2}\Big[\Big(\dfrac{n}{2}\Big)^{th} \text{ term} + \Big(\dfrac{n}{2} + 1\Big)^{th} \text{term}\Big]$

So, statement 2 is true.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Statement 1: The mean of 100 observation is 50 and one of these observation is increased by 150, the sum of resulting observation is 100 x 50 + 150.

Statement 2: The sum of resulting observation = 100 x 50 + 100.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given,

Number of observations = 100

Mean = 50

By formula,

Mean = $\dfrac{\text{Sum of observations}}{\text{Number of observation}}$

$\Rightarrow 50 = \dfrac{\text{Sum of observations}}{100}\\[1em] \Rightarrow \text{Sum of observations} = 50 \times 100.$

One of the observation is increased by 150, then sum of the observation = Original sum + 150

⇒ Sum of the observation = 50 x 100 + 150

∴ Statement 1 is true, and statement 2 is false.

Hence, option 3 is the correct option.

Assertion (A): The mean of x₁, x₂ and x₃ is m. Then the value of (x₁ - m) + (x₂ - m) + (x₃ - m) = 0.

Reason (R): x₁ + x₂ + x₃ = 3m

⇒ (x₁ - m) + (x₂ - m) + (x₃ - m) = (x₁ + x₂ + x₃) - 3m

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given,

Observations = x₁, x₂ and x₃

Mean = m

Number of observations = 3

By formula,

Mean = $\dfrac{\text{Sum of observations}}{\text{Number of observation}}$

$\Rightarrow m = \dfrac{x_1 + x_2 + x_3}{3}\\[1em] \Rightarrow 3m = x_1 + x_2 + x_3\\[1em] \Rightarrow x_1 + x_2 + x_3 - 3m = 0\\[1em] \Rightarrow (x_1 - m) + (x_2 - m) + (x_3 - m) = 0\\[1em]$

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Assertion (A): Mean of n observations is x and mean of another set of n observations is y, the combined mean of all the observations is $\dfrac{x + y}{2}$

Reason (R): Total of all the observations = nx + ny

∴ Mean of all the observations = $\dfrac{nx + ny}{2n}$

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given,

For 1st set :

Mean = x

Number of observations = n

By formula,

Mean = $\dfrac{\text{Sum of observations}}{\text{Number of observation}}$

Substituting values we get :

$\Rightarrow x = \dfrac{\text{Sum of observations}}{n} \\[1em] \Rightarrow \text{Sum of observations} = x \times n \\[1em] \Rightarrow \text{Sum of observations} = xn.$

For 2nd set :

Mean = y

Number of observations = n

Substituting values we get :

$\Rightarrow y = \dfrac{\text{Sum of observations}}{n} \\[1em] \Rightarrow \text{Sum of observations} = y \times n \\[1em] \Rightarrow \text{Sum of observations} = yn. \\[1em] \text{Mean of all observations } = \dfrac{\text{Sum of observations of 1st set + Sum of observations of 2nd set}}{\text{Total number of observations}}\\[1em] = \dfrac{nx + ny}{n + n}\\[1em] = \dfrac{n(x + y)}{2n}\\[1em] = \dfrac{x + y}{2}.$

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

The mean of 100 observations is 40. It is found that an observation 53 was misread as 83. Find the correct mean.

Answer

Given:

Number of observations = 100

Mean = 40

Mean = $\dfrac{\text{Sum of all observations}}{\text{Number of all observations}}$

⇒ 40 = $\dfrac{\text{Sum of all observations }}{100}$

⇒ Sum of all observations = 40 x 100

⇒ Sum of all observations = 4,000

Correct sum of observations = Incorrect sum - incorrect observation + correct observation

= 4,000 - 83 + 53

= 3,970

Correct Mean = $\dfrac{3,970}{100}$

= 39.7

Hence, the correct mean is 39.7.

The mean of 200 items was 50. Later on, it was discovered that two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

Answer

Given:

Number of observations = 200

Mean = 50

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

⇒ 50 = $\dfrac{\text{Sum of all observations }}{200}$

⇒ Sum of all observations = 50 x 200

⇒ Sum of all observations = 10,000

Correct sum of observations = Incorrect sum - incorrect observations + correct observations

= 10,000 - (92 + 8) + (192 + 88)

= 10,000 - 100 + 280

= 10,180

Correct Mean = $\dfrac{10,180}{200}$

= 50.9

Hence, the correct mean is 50.9.

Find the mean of 75 numbers, if the mean of 45 of them is 18 and the mean of the remaining ones is 13.

Answer

Mean of 45 numbers = 18

⇒ Total of 45 numbers = 45 x 18 = 810

Mean of remaining (75 - 45) 30 numbers = 13

⇒ Total of 30 numbers = 13 x 30 = 390

∴ Total of all the 75 numbers = 810 + 390 = 1,200

⇒ Mean of all the 75 numbers = $\dfrac{1200}{75}$ = 16

Hence, the mean of 75 numbers is 16.

The mean weight of 120 students of a school is 52.75 kg. If the mean weight of 50 of them is 51 kg, find the mean weight of the remaining students.

Answer

Mean weight of 120 students = 52.75 kg

Let a be the sum of weights of 120 students.

Mean weight of 120 students = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

⇒ 52.75 = $\dfrac{a}{120}$

⇒ a = 120 x 52.75

⇒ a = 6,330

Mean weight of 50 students = 51 kg

Let b be the sum of weights of 50 students.

⇒ 51 = $\dfrac{b}{50}$

⇒ b = 50 x 51

⇒ b = 2,550

Sum of weights of remaining (120 - 50) 70 students = 6,330 - 2,550 = 3,780

Mean weight of remaining 70 students = $\dfrac{3780}{70}$

= 54

Hence, the mean weight of remaining 70 students is 54 kg.

The mean marks (out of 100) of boys and girls in an examination are 70 and 73 respectively. If the mean marks of all the students in that examination is 72.25, find the ratio of the number of boys to the number of girls.

Answer

Given,

Mean marks of boys in the examination = 70

Mean marks of girls in the examination = 73

Let the number of boys and girls be a and b respectively.

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

Mean marks of boys = $\dfrac{\text{Sum of all marks of boys}}{\text{Number of all boys}}$

⇒ 70 = $\dfrac{\text{Sum of all marks of boys}}{a}$

⇒ Sum of all marks of boys = 70a

Mean marks of girls = $\dfrac{\text{Sum of all marks of girls}}{\text{Number of all girls}}$

⇒ 73 = $\dfrac{\text{Sum of all marks of girls}}{b}$

⇒ Sum of all marks of girls = 73b

Mean of all students = $\dfrac{\text{Sum of marks of all students}}{\text{Number of all students}}$

⇒ 72.25 = $\dfrac{70a + 73b}{a + b}$

⇒ 72.25(a + b) = 70a + 73b

⇒ 72.25a + 72.25b = 70a + 73b

⇒ 72.25a - 70a = 73b - 72.25b

⇒ 2.25a = 0.75b

⇒ $\dfrac{a}{b} = \dfrac{0.75}{2.25} = \dfrac{1}{3}$.

Hence, the ratio of the number of boys to the number of girls is 1 : 3.

Find x, if 9, x, 14, 18, x, x, 8, 10 and 4 have a mean of 11.

Answer

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations}}$

⇒ 11 = $\dfrac{9 + x + 14 + 18 + x + x + 8 + 10 + 4}{9}$

⇒ 11 = $\dfrac{63 + 3x}{9}$

⇒ 11 x 9 = 63 + 3x

⇒ 99 = 63 + 3x

⇒ 3x = 99 - 63

⇒ 3x = 36

⇒ x = $\dfrac{36}{3}$

⇒ x = 12

Hence, the value of x is 12.

In a series of tests, A appeared for 8 tests. Each test was marked out of 30 and averages 25. However, while checking his files, A could only find 7 of the 8 tests. For these he scored 29, 26, 18, 20, 27, 24 and 29. Determine how many marks he scored for the eighth test.

Answer

Total number of tests = 8

The average score of A = 25

Let the score of the 8^th test be a.

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

⇒ 25 = $\dfrac{29 + 26 + 18 + 20 + 27 + 24 + 29 + a}{8}$

⇒ 25 = $\dfrac{173 + a}{8}$

⇒ 173 + a = 25 x 8

⇒ 173 + a = 200

⇒ a = 200 - 173

⇒ a = 27

Hence, A scored 27 marks in the eighth test.

Find the mean of 8, 12, 16, 22, 10 and 4. Find the resulting mean, if each of the observations, given above, be :

(i) multiplied by 3.

(ii) divided by 2.

(iii) multiplied by 3 and then divided by 2.

(iv) increased by 25 %.

(v) decreased by 40 %.

Answer

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

= $\dfrac{8 + 12 + 16 + 22 + 10 + 4}{6}$

= $\dfrac{72}{6}$

= 12

(i) According to property 4, if each observation is multiplied by a quantity a, then the mean is also multiplied by the same quantity a.

If each observation is multiplied by 3, then the mean is also multiplied by 3 ( 12 x 3 = 36).

Hence, the mean of the new observations is 36.

(ii) According to property 5, if each observation is divided by a quantity a, then the mean is also divided by the same quantity a.

If each observation of the data is divided by 2, then the mean is also divided by 2 $\Big(\dfrac{12}{2} = 6\Big)$.

Hence, the mean of the new observations is 6.

(iii) If each observation of data is multiplied by 3, then the mean is also multiplied by 3 ( 12 x 3 = 36).

Similarly, if each observation of the data is divided by 2, then the mean is also divided by 2 $\Big(\dfrac{36}{2} = 18\Big)$.

Hence, the mean of the new observations is 18.

(iv) According to property 2, if each observation is increased by a quantity a, then the mean is also increased by the same quantity a.

If each observation is increased by 25 %, then the mean is also increased by 25 %.

= 12 + $\dfrac{25}{100} \times 12$

= 12 + 3

= 15

Hence, the mean of the new observations is 15.

(v) According to property 3, if each observation is decreased by a quantity a, then the mean is also decreased by the same quantity a.

If each observation is decreased by 40 %, then the mean is also decreased by 40 %.

= 12 - $\dfrac{40}{100} \times 12$

= 12 - 4.8

= 7.2

Hence, the mean of the new observations is 7.2.

The mean of 18, 24, 15, 2x + 1 and 12 is 21. Find the value of x.

Answer

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

⇒ 21 = $\dfrac{18 + 24 + 15 + 2x + 1 + 12}{5}$

⇒ 21 = $\dfrac{70 + 2x}{5}$

⇒ 70 + 2x = 21 x 5

⇒ 70 + 2x = 105

⇒ 2x = 105 - 70

⇒ 2x = 35

⇒ x = $\dfrac{35}{2}$

⇒ x = 17.5

Hence, the value of x is 17.5.

The mean of 6 numbers is 42. If one number is excluded, the mean of remaining numbers is 45. Find the excluded number.

Answer

Given:

Number of observations = 6

Mean = 42

⇒ Sum of all 5 observation = 6 x 42 = 252

On excluding an observation, the mean of the remaining 5 observation = 45

∵ Sum of all remaining 5 observation = 5 x 45 = 225

⇒ Excluded observation = Sum of all 6 observations - Sum of all remaining 5 observations

= 252 - 225

= 27

Hence, the excluded number is 27.

The mean of 10 numbers is 24. If one more number is included, the new mean is 25. Find the included number.

Answer

Given:

Number of observations = 10

Mean = 24

⇒ Sum of all 10 observations = 10 x 24 = 240

On including an observation, the mean of the 11 observations = 25

∵ Sum of all 11 observations = 11 x 25 = 275

⇒ Included observation = Sum of all 11 observations - Sum of all 10 observations

= 275 - 240

= 35

Hence, the included number is 35.

The following observations have been arranged in ascending order. If the median of the data is 78, find the value of x.

44, 47, 63, 65, x + 13, 87, 93, 99, 110.

Answer

Number of observations, n = 9 (odd)

Median = $\Big[\dfrac{n+1}{2}\Big]^{th}$ term

⇒ 78 = $\Big[\dfrac{9+1}{2}\Big]^{th}$ term

⇒ 78 = $\Big[\dfrac{10}{2}\Big]^{th}$ term

⇒ 78 = $5^{th}$ term

⇒ 78 = x + 13

⇒ x = 78 - 13

⇒ x = 65

Hence, the value of x is 65.

The following observations have been arranged in ascending order. If the median of these observations is 58, find the value of x.

24, 27, 43, 48, x - 1, x + 3, 68, 73, 80, 90.

Answer

Number of observations, n = 10 (even)

Median = $\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{n}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{n}{2} + 1\Big)^{th}\Big]$ term

⇒ 58 = $\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{10}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{10}{2} + 1\Big)^{th}\Big]$ term

⇒ 58 = $\dfrac{1}{2}\Big[\text{the value of} (5)^{th} + \text{the value of} (5 + 1)^{th}\Big]$ term

⇒ 58 = $\dfrac{1}{2}\Big[(x - 1) + (x + 3)\Big]$

⇒ 58 = $\dfrac{1}{2}\Big[x - 1 + x + 3\Big]$

⇒ 58 = $\dfrac{1}{2}\Big[2x + 2\Big]$

⇒ 58 = x + 1

⇒ x = 58 - 1

⇒ x = 57

Hence, the value of x is 57.

Find the mean of the following data :

30, 32, 24, 34, 26, 28, 30, 35, 33, 25

(i) Show that the sum of the deviations of all the given observations from the mean is zero.

(ii) Find the median of the given data.

Answer

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

= $\dfrac{30 + 32 + 24 + 34 + 26 + 28 + 30 + 35 + 33 + 25}{10}$

= $\dfrac{297}{10}$

= 29.7

Hence, the mean is 29.7.

(i) $\overline{x}$ = 29.7

$∑(x - \overline{x})$ = (-0.3) + (-2.3) + 5.7 + (-4.3) + 3.7 + 1.7 + (-0.3) + (-5.3) + (-3.3) + 4.7

= -0.3 - 2.3 + 5.7 - 4.3 + 3.7 + 1.7 - 0.3 - 5.3 - 3.3 + 4.7

= 0

Hence, the sum of the deviations of all the given observations from the mean is zero.

(ii) On arranging the given set of data in ascending order, we get :

24, 25, 26, 28, 30, 30, 32, 33, 34, 35

Number of observations, n = 10 (even)

Median = $\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{n}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{n}{2} + 1\Big)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{10}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{10}{2} + 1\Big)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[\text{the value of} (5)^{th} + \text{the value of} (5 + 1)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[30 + 30\Big]$

= $\dfrac{1}{2}\Big[60\Big]$

= 30

Hence, the median is 30.

Find the mean and median of the data :

35, 48, 92, 76, 64, 52, 51, 63 and 71.

If 51 is replaced by 66, what will be the new median ?

Answer

Mean = $\dfrac{\text{Sum of all observations}}{\text{Number of all observations}}$

= $\dfrac{35 + 48 + 92 + 76 + 64 + 52 + 51 + 63 + 71}{9}$

= $\dfrac{552}{9}$

= 61.33

Hence, the mean is 61.33.

On arranging the given set of data in ascending order, we get :

35, 48, 51, 52, 63, 64, 71, 76, 92

Number of observations, n = 9 (odd)

Median = $\Big[\dfrac{n+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{9+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{10}{2}\Big]^{th}$ term

= $5^{th}$ term

= 63

Hence, the median is 63.

When 51 is replaced by 66, and the given set of data is arranged in ascending order, we get :

35, 48, 52, 63, 64, 66, 71, 76, 92

Number of observations, n = 9 (odd)

Median = $\Big[\dfrac{n+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{9+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{10}{2}\Big]^{th}$ term

= $5^{th}$ term

= 64

Hence, the new median is 64.

The mean of x, x + 2, x + 4, x + 6 and x + 8 is 11, find the mean of the first three observations.

Answer

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

⇒ 11 = $\dfrac{x + (x + 2) + (x + 4) + (x + 6) + (x + 8)}{5}$

⇒ 11 = $\dfrac{x + x + 2 + x + 4 + x + 6 + x + 8}{5}$

⇒ 11 = $\dfrac{5x + 20}{5}$

⇒ 11 x 5 = 5x + 20

⇒ 55 = 5x + 20

⇒ 5x = 55 - 20

⇒ 5x = 35

⇒ x = 7

So, the observations are x, x + 2, x + 4, x + 6 and x + 8

= 7, 9, 11, 13, 15

Mean of the first three observations = $\dfrac{7 + 9 + 11}{3}$

= $\dfrac{27}{3}$

= 9

Hence, the mean of the first three observations is 9.

Find the mean and median of all the positive factors of 72.

Answer

The positive factors are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

= $\dfrac{1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 36 + 72}{12}$

= $\dfrac{195}{12}$

= 16.25

Hence, the mean is 16.25.

On arranging the given set of data in ascending order, we get :

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

Number of observations, n = 12 (even)

Median = $\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{n}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{n}{2} + 1\Big)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{12}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{12}{2} + 1\Big)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[\text{the value of} (6)^{th} + \text{the value of} (6 + 1)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[8 + 9\Big]$

= $\dfrac{1}{2}\Big[17\Big]$

= 8.5

Hence, the median is 8.5.

The mean weight of 60 students in a class is 40 kg. The mean weight of boys is 50 kg while that of girls is 30 kg. Find the number of boys and girls in the class.

Answer

Given:

Mean weight of boys = 50

Mean weight of girls = 30

Let the number of boys and girls be a and b respectively.

Mean weight of boys = $\dfrac{\text{Sum of all weights of boys}}{\text{Number of all boys}}$

⇒ 50 = $\dfrac{\text{Sum of all weights of boys}}{a}$

⇒ Sum of all weights of boys = 50a

Mean weight of girls = $\dfrac{\text{Sum of all weights of girls}}{\text{Number of all girls}}$

⇒ 30 = $\dfrac{\text{Sum of all weights of girls}}{b}$

⇒ Sum of all weights of girls = 30b

Mean weight of 60 students = $\dfrac{\text{Sum of all weights of students}}{\text{Number of all students}}$

⇒ 40 = $\dfrac{50a + 30b}{a + b}$

⇒ 40(a + b) = 50a + 30b

⇒ 40a + 40b = 50a + 30b

⇒ 50a - 40a = 40b - 30b

⇒ 10a = 10b

⇒ a = b

Total students = 60

⇒ a + b = 60

⇒ a + a = 60

⇒ 2a = 60

⇒ a = $\dfrac{60}{2}$

⇒ a = 30

⇒ b = 30

Hence, the number of boys is 30 and the number of girls is 30.

The average of n numbers x₁, x₂, x₃ ..............., x_n is A. If x₁ is replaced by (x + a)x₁, x₂ is replaced by (x + a)x₂ and so on. Find the new average.

Answer

Average of n numbers = A

⇒ A = $\dfrac{x_1 + x_2 + x_3 + ....... + x_n}{n}$

⇒ $x_1 + x_2 + x_3 + ....... + x_n = nA$ ..........(1)

New average = $\dfrac{(x + a)x_1 + (x + a)x_2 + (x + a)x_3 + ....... + (x + a)x_n}{n}$

= $\dfrac{(x + a)(x_1 + x_2 + x_3 + ....... + x_n)}{n}$

= $\dfrac{(x + a)(nA)}{n}$ [∵Using equation (1)]

= $(x + a)A$

Hence, new average is (x + a)A.

The heights (in cm) of the volley-ball players from team A and team B were recorded as :

Team A : 180, 178, 176, 181, 190, 175, 187

Team B : 174, 175, 190, 179, 178, 185, 177

Which team had the greater average height ? Find the median of team A and team B.

Answer

Total number of players in each team = 7

Mean = $\dfrac{\text{Sum of the height of players of the team A}}{\text{Number of all players of team A}}$

Mean height of team A = $\dfrac{180 + 178 + 176 + 181 + 190 + 175 + 187}{7}$

= $\dfrac{1,267}{7}$

= 181 cm

Mean = $\dfrac{\text{Sum of the height of players of the team B}}{\text{Number of all players of team B}}$

Mean height of team B = $\dfrac{174 + 175 + 190 + 179 + 178 + 185 + 177}{7}$

= $\dfrac{1,258}{7}$

= 179.7 cm

Median of team A,

On arranging the given set of data in ascending order, we get :

175, 176, 178, 180, 181, 187, 190

Number of observations, n = 7 (odd)

Median = $\Big[\dfrac{n+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{7+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{8}{2}\Big]^{th}$ term

= $4^{th}$ term

= 180 cm

Median of team B,

On arranging the given set of data in ascending order, we get :

174, 175, 177, 178, 179, 185, 190

Number of observations, n = 7 (odd)

Median = $\Big[\dfrac{n+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{7+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{8}{2}\Big]^{th}$ term

= $4^{th}$ term

= 178 cm

Hence, team A has greater average height. Median of team A is 180 cm and team B is 178 cm.

The mean of the following arrayed data 5, 8, (3x - 1), (4x + 1), (3x + 7) is equal to its median. Find the value of x.

Answer

Given,

5, 8, (3x - 1), (4x + 1), (3x + 7)

$\text{Mean} = \dfrac{\text{Sum of observations}}{\text{Number of observations}} \\[1em] = \dfrac{5 + 8 + 3x - 1 + 4x + 1 + 3x + 7}{5} \\[1em] = \dfrac{10x + 20}{5} \\[1em] = 2x + 4$

Since there are 5 observations, the median is $\dfrac{5 + 1}{2}\text{th term}$ = 3rd term.

So, Median = 3x - 1

Since, Mean = Median

⇒ 2x + 4 = 3x - 1

⇒ 3x - 2x = 4 + 1

⇒ x = 5.

Hence, x = 5.