Data Handling - Exercise 24(A)

Find the arithmetic mean of first five prime numbers.

Answer

The first five prime numbers are 2, 3, 5, 7 and 11.

Sum of all observations = 2 + 3 + 5 + 7 + 11 = 28

Number of observations = 5

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

$= \dfrac{28}{5} \\[1em] = 5.6$

Hence, the arithmetic mean of first five prime numbers is 5.6.

The marks obtained by 12 students in an examination (out of 50) are given below:

18, 35, 2, 27, 40, 0, 21, 33, 27, 8, 36, 23

Find the mean marks.

Answer

Sum of all observations = 18 + 35 + 2 + 27 + 40 + 0 + 21 + 33 + 27 + 8 + 36 + 23 = 270

Number of observations = 12

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

$= \dfrac{270}{12} \\[1em] = 22.5$

Hence, the mean marks is 22.5.

In a one-day cricket match, the runs scored by the players of a team are

6, 10, 16, 20, 8, 19, 30, 57, 2, 0, 8.

Find the mean score.

Answer

Sum of all observations = 6 + 10 + 16 + 20 + 8 + 19 + 30 + 57 + 2 + 0 + 8 = 176

Number of observations = 11

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

$= \dfrac{176}{11} \\[1em] = 16$

Hence, the mean score is 16.

The following are the ages (in years) of 12 teachers in a school :

36, 44, 39, 46, 35, 53, 38, 42, 55, 45, 49, 40

Arrange the above data in an ascending order and answer the questions given below:

(i) What is the age of the eldest teacher in the school?

(ii) What is the age of the youngest teacher in the school?

(iii) What is the range of the ages of the teachers in the school?

(iv) What is the mean age of the teachers in the school?

Answer

Arranging the given data in ascending order, we get :

35, 36, 38, 39, 40, 42, 44, 45, 46, 49, 53, 55

(i) The last number in our sorted list is the maximum value i.e., 55.

Hence, the age of the eldest teacher is 55 years.

(ii) The first number in our sorted list is the minimum value i.e., 35.

Hence, the age of the youngest teacher is 35 years.

(iii) Range = Highest observation - Lowest observation

= 55 years - 35 years

= 20 years.

Hence, the range of the ages of the teachers is 20 years.

(iv) Sum of all observations = 35 + 36 + 38 + 39 + 40 + 42 + 44 + 45 + 46 + 49 + 53 + 55 = 522

Number of observations = 12

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

$= \dfrac{522}{12} \\[1em] = 43.5$

Hence, the mean age of the teachers is 43.5 years.

The daily wages (in ₹) of 15 workers in a factory are given below :

195, 185, 145, 155, 135, 180, 175, 200, 150, 125, 190, 180, 170, 175, 190

(i) Find the mean daily wage.

(ii) Find the range of the data.

Answer

(i) Sum of all observations = 195 + 185 + 145 + 155 + 135 + 180 + 175 + 200 + 150 + 125 + 190 + 180 + 170 + 175 + 190 = 2550

Number of observations = 15

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

$= ₹ \left(\dfrac{2550}{15}\right) \\[1em] = ₹ 170$

Hence, the mean daily wage is ₹ 170.

(ii) Range = Highest observation - Lowest observation

= ₹ 200 - ₹ 125

= ₹ 75

Hence, the range of the data is ₹ 75.

The maximum daily temperatures (in °C) of a city during a week are given below :

28.9, 32.6, 24.6, 26.1, 29.2, 30 and 27.4

(i) Find the mean temperature.

(ii) Find the range of the data.

Answer

(i) Sum of all observations = 28.9 + 32.6 + 24.6 + 26.1 + 29.2 + 30 + 27.4 = 198.8

Number of observations = 7

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

$= \left(\dfrac{198.8}{7}\right) \degree C \\[1em] = 28.4 \degree C$

Hence, the mean temperature is 28.4 °C.

(ii) Range = Highest observation - Lowest observation

= 32.6 °C - 24.6 °C

= 8 °C

Hence, the range of the data is 8 °C.

If the mean of 4, 6, x, 9, 10, 5 is 7, find the value of x.

Answer

Given:

Observations are 4, 6, x, 9, 10, 5

Mean = 7

Sum of all observations = 4 + 6 + x + 9 + 10 + 5 = 34 + x

Number of observations = 6

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

$⇒ 7 = \dfrac{34 + x}{6} \\[1em] ⇒ 7 \times 6 = 34 + x \\[1em] ⇒ 42 = 34 + x \\[1em] ⇒ x = 42 - 34 \\[1em] ⇒ x = 8$

Hence, the value of x is 8.

The number of children in 25 families are given below :

2, 2, 1, 4, 2, 3, 2, 2, 1, 1, 1, 3, 4, 3, 2, 0, 1, 3, 3, 1, 2, 4, 2, 0, 2

Represent the above data in the form of frequency distribution.

Answer

Arranging the given data in ascending order, we get :

0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4

The frequency distribution table is :

A dice was thrown 30 times and the following outcomes were noted :

1, 3, 3, 2, 5, 4, 4, 6, 1, 2, 2, 3, 4, 6, 2, 3, 3, 4, 1, 2, 3, 3, 4, 5, 6, 3, 2, 1, 3, 4

Represent the above data in the form of frequency distribution.

Answer

Arranging the given data in ascending order, we get :

1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6

The frequency distribution table is :

Find the mean weight of 50 boys from the following data :

Answer

We calculate the mean as follows:

Mean = $\dfrac{∑f_ix_i}{∑f_i}$

= $\dfrac{2730}{50}$ kg

= 54.6 kg

Hence, the mean weight is 54.6 kg.

The heights (in cm) of 90 plants in a garden are given below :

Find the mean height.

Answer

We calculate the mean as follows:

Mean = $\dfrac{∑f_ix_i}{∑f_i}$

= $\dfrac{5634}{90}$ cm

= 62.6 cm

Hence, the mean height is 62.6 cm.

Find the mean height of 65 boys from the following data :

Answer

We calculate the mean as follows:

Mean = $\dfrac{∑f_ix_i}{∑f_i}$

= $\dfrac{9542}{65}$ cm

= 146.8 cm

Hence, the mean height is 146.8 cm.

If the mean of the following frequency distribution is 15, find the value of p.

Answer

For calculating the mean, we prepare the following table:

Mean = $\dfrac{∑f_ix_i}{∑f_i}$

⇒ 15 = $\dfrac{1356 + 12p}{88 + p}$

⇒ 15(88 + p) = 1356 + 12p

⇒ 1320 + 15p = 1356 + 12p

⇒ 15p - 12p = 1356 - 1320

⇒ 3p = 36

⇒ p = $\dfrac{36}{3}$

⇒ p = 12

Hence, the value of p is 12.