Mean & Median of Ungrouped Data & Frequency Polygon — Exercise 16(A)

The weight of 7 boys in a group are 52 kg, 57 kg, 55 kg, 60 kg, 54 kg, 59 kg and 55 kg. Find the mean weight of the group.

Answer

Weights of 7 boys (in kg) :

52, 57, 55, 60, 54, 59, 55

Number of boys (n) = 7

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

Sum of observations = 52 + 57 + 55 + 60 + 54 + 59 + 55 = 392

Total number of observations = 7

Mean = $\dfrac{392}{7}$ = 56.

Hence, mean weight of the group is 56 kg.

The marks obtained by 7 students in a group are 340, 180, 260, 164, 56, 275 and 307 respectively. Find the mean marks per student.

Answer

Marks of 7 students:

340, 180, 260, 164, 56, 275, 307

Number of students (n) = 7

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

Sum of observations = 340 + 180 + 260 + 164 + 56 + 275 + 307 = 1582

Total number of observations = 7

Mean = $\dfrac{ 1582}{7}$ = 226.

Hence, mean marks per student is 226.

Find the mean of first six prime numbers.

Answer

First six prime numbers are:

2, 3, 5, 7, 11, 13

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

Sum of prime numbers = 2 + 3 + 5 + 7 + 11 + 13 = 41

Mean = $\dfrac{41}{6} \approx$ 6.83

Hence, mean of first six prime numbers is 6.83.

Find the mean of the first ten odd numbers.

Answer

First ten odd numbers are:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Sum of observations = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

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

= $\dfrac{100}{10}$ = 10.

Hence, mean of first ten odd numbers is 10.

Find the mean of all the factors of 20.

Answer

Factors of 20 are the numbers that divide 20 exactly.

The factors are :

1, 2, 4, 5, 10, 20

Total number of observations = 6

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

= $\dfrac{1 + 2 + 4 + 5 + 10 + 20}{6}$

= $\dfrac{42}{6}$ = 7.

Hence, mean of all the factors of 20 is 7.

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

Find the mean temperature.

Answer

We know that,

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

= $\dfrac{35.2 + 31.1 + 27.6 + 31.8 + 29.3 + 23.8}{6}$

= $\dfrac{178.8}{6}$

= 29.8 °F

Hence, mean temperature = 29.8 °F.

If the mean of 6, 8, 9, x, 13 is 10, find the value of x.

Answer

Numbers: 6, 8, 9, x, 13

Mean = 10

Number of observations (n) = 5

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

⇒ 10 = $\dfrac{6 + 8 + 9 + x + 13}{5}$

⇒ 50 = 36 + x

⇒ x = 50 - 36 = 14.

Hence, x = 14.

The mean of the heights of 6 girls is 148 cm. If the individual heights of five of them are 142 cm, 154 cm, 146 cm, 145 cm and 150 cm, find the height of the sixth girl.

Answer

Mean height of 6 girls = 148 cm

Number of girls (n) = 6

Heights of 5 girls:

142 cm, 154 cm, 146 cm, 145 cm, 150 cm

Let height of the sixth girl be x cm.

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

⇒ 148 = $\dfrac{142 + 154 + 146 + 145 + 150 + x}{6}$

⇒ 148 × 6 = 737 + x

⇒ 888 = 737 + x

⇒ x = 888 - 737 = 151.

Hence, height of the sixth girl = 151 cm.

The following table shows the weights (in kg) of 15 workers in a factory :

Calculate the mean weight.

Answer

$\text{Mean} = \dfrac{\sum fx}{\sum f}$

where

f = frequency (Number of workers)

x = observation (weight)

So, ∑fx = 975 and ∑f = 15

Therefore,

$\text{Mean} = \dfrac{\sum fx}{\sum f}$

= $\dfrac{975}{15}$ = 65.

Hence, mean weight = 65 kg.

Find the mean of daily wages of 60 workers in a factory as per data given below:

Answer

$\text{Mean} = \dfrac{\sum fx}{\sum f}$

where

f = frequency (Number of workers)

x = observation (Daily wages)

so, ∑fx = 7110 and ∑f = 60

Mean = $\dfrac{7110}{60}$ = 118.5.

Hence, mean of daily wages of 60 workers is ₹118.5

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

Calculate the mean height.

Answer

$\text{Mean} = \dfrac{\sum fx}{\sum f}$

where

f = frequency (Number of plants)

x = observation (Height)

∑fx = 5634 and ∑f = 90

Mean = $\dfrac{5634}{90}$ = 62.6

Hence, mean height = 62.6 cm.

The mean of the following data is 21.6. Find the value of p.

Answer

Mean = $\dfrac{\sum f_ix_i}{\sum f_i}$

∑f_i = 5 + 4 + p + 6 + 4 + 6 = 25 + p

∑f_ix_i​ = 30 + 48 + 18p + 144 + 120 + 216 = 558 + 18p

⇒ 21.6 = $\dfrac{558+18p}{25+p}$

⇒ 21.6(25 + p) = 558 + 18p

⇒ 540 + 21.6p = 558 + 18p

⇒ 21.6p − 18p = 558 − 540

⇒ 3.6p = 18

⇒ p = $\dfrac{18}{3.6}$ = 5

Hence, p = 5.

If the mean of the following data is 18.75, find the value of p.

Answer

Mean = 18.75

∑f_i​ = 5 + 10 + 7 + 8 + 2 = 32

∑f_ix_i​ = 50 + 150 + 7p + 200 + 60 = 460 + 7p

Mean = $\dfrac{\sum f_ix_i}{\sum f_i}$

⇒ 18.75 = $\dfrac{460 + 7p}{32}$

⇒ 600 = 460 + 7p

⇒ 7p = 600 - 460

⇒ 7p = 140

⇒ p = $\dfrac{140}{7}$ = 20.

Hence, p = 20.

The mean age of a group of 40 students is 17.45 years. Find the missing frequencies.

Answer

Let missing frequencies be:

Frequency at age 16 = a

Frequency at age 19 = b

By formula,

Mean = $\dfrac{\sum f_ix_i}{\sum f_i}$

where

x_i = Age (in years)

f_i = Number of students

∑f_i = 3 + a + 9 + 11 + b + 3 = 40

⇒ a + b + 26 = 40

⇒ a + b = 14

⇒ b = 14 - a ........(1)

Mean = $\dfrac{\sum f_ix_i}{\sum f_i}$

⇒ 17.45 = $\dfrac{45 + 16a + 153 + 198 + 19b + 60}{40}$

⇒ 17.45 × 40 = 456 + 16a + 19b

⇒ 698 = 456 + 16a + 19b

⇒ 242 = 16a + 19b ..........(2)

Substitute equation 1 in equation 2

⇒ 16a + 19(14 − a) = 242

⇒ 16a + 266 − 19a = 242

⇒ −3a + 266 = 242

⇒ −3a = −24

⇒ a = 8.

⇒ Put a = 8 in equation 1

⇒ b = 14 - 8

⇒ b = 6.

Hence, the missing frequencies are 8 & 6.

Using the assumed mean method, calculate the mean weekly wage from the following frequency distribution :

Answer

By using assumed mean method,

Let A = 1100

By formula,

Mean = A + $\dfrac{\sum f_id_i}{\sum f_i}$

= 1100 + $\dfrac{5850}{110}$

= 1100 + 53.18

= 1153.18.

Hence, mean weekly wage is ₹ 1153.18.

Using the step-deviation method, find the mean from the following data :

Answer

We construct the following table, taking assumed mean A = 21

Here c (width of the class) = 1

Mean = A + c × $\dfrac{\sum f_iu_i}{\sum f_i}$

= 21 + 1 × $\dfrac{-840}{2200}$

= 21 + $\dfrac{-840}{2200}$

= 21 - 0.382 = 20.62.

Hence, mean of following distribution is 20.62.