Class 10 - Chapter 13: Statistics

Exercise 13.3

Q1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Mean

Mean = Σfx / Σf

= 9320 / 68

= 137.06

Mean = 137.06 units

Median

Total frequency (N) = 68

N/2 = 34

Median Class = 125-145

l = 125, h = 20

f = 20

cf = 22

Median = l + [(N/2 − cf)/f] × h

= 125 + [(34 − 22)/20] × 20

= 125 + 12

= 137

Median = 137 units

Mode

Modal Class = 125-145

l = 125, h = 20

f₁ = 20, f₀ = 13, f₂ = 14

Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h

= 125 + [(20 − 13)/(40 − 13 − 14)] × 20

= 125 + (7/13) × 20

= 125 + 10.77

= 135.77

Mode = 135.77 units

Comparison

Mode ≈ 135.77

Median = 137

Mean ≈ 137.06

These three values are very close to each other, showing that the distribution is nearly symmetrical.

Q2. If the median of the distribution given below is 28.5, find the values of x and y.

Total frequency:

5 + x + 20 + 15 + y + 5 = 60

x + y = 15 .......... (1)

N = 60

N/2 = 30

Median = 28.5

Median Class = 20-30

l = 20, h = 10

f = 20

cf = 5 + x

Using:

Median = l + [(N/2 − cf)/f] × h

28.5 = 20 + [(30 − (5 + x))/20] × 10

8.5 = (25 − x)/2

17 = 25 − x

x = 8

From equation (1):

x + y = 15

8 + y = 15

y = 7

x = 8 and y = 7

Q3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age.

First convert the less-than cumulative frequency table into a frequency distribution table.

Total frequency (N) = 100

N/2 = 50

The cumulative frequency just greater than 50 is 78.

Therefore, the median class is:

35 - 40

For the median class:

l = 35

cf = 45

f = 33

h = 5

Using the formula:

Median = l + [(N/2 − cf)/f] × h

= 35 + [(50 − 45)/33] × 5

= 35 + (5/33) × 5

= 35 + 25/33

= 35 + 0.758

= 35.76

Median Age = 35.76 years

Q4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre. Find the median length of the leaves.

Since the measurements are recorded to the nearest millimetre, convert the classes into continuous intervals.

Total number of leaves (N) = 40

N/2 = 40/2 = 20

The cumulative frequency just greater than 20 is 29.

Therefore, the median class is:

144.5 - 153.5

For the median class:

l = 144.5

f = 12

cf = 17

h = 9

Using the median formula:

Median = l + [(N/2 − cf) / f] × h

= 144.5 + [(20 − 17) / 12] × 9

= 144.5 + (3/12) × 9

= 144.5 + 2.25

= 146.75

Median Length = 146.75 mm

Q5. The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp.

Total frequency (N) = 400

N/2 = 200

The cumulative frequency just greater than 200 is 216.

Therefore, the median class is:

3000 - 3500

l = 3000, h = 500

f = 86

cf = 130

Median = l + [(N/2 − cf) / f] × h

= 3000 + [(200 − 130) / 86] × 500

= 3000 + (70/86) × 500

= 3000 + 406.98

= 3406.98

Median life time = 3406.98 hours

Q6. Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

Mean

Mean = Σfx / Σf

= 832 / 100

= 8.32

Mean = 8.32 letters

Median

N = 100

N/2 = 50

Median class = 7-10

l = 7

f = 40

cf = 36

h = 3

Median = l + [(N/2 − cf) / f] × h

= 7 + [(50 − 36)/40] × 3

= 7 + 1.05

= 8.05

Median = 8.05 letters

Mode

Modal Class = 7-10

l = 7

h = 3

f₁ = 40

f₀ = 30

f₂ = 16

Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h

= 7 + [(40 − 30)/(80 − 30 − 16)] × 3

= 7 + (10/34) × 3

= 7 + 0.88

= 7.88

Mode = 7.88 letters

Q7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

N = 30

N/2 = 15

Median Class = 55-60

l = 55

f = 6

cf = 13

h = 5

Median = l + [(N/2 − cf) / f] × h

= 55 + [(15 − 13)/6] × 5

= 55 + 1.67

= 56.67

Median Weight = 56.67 kg