Class 10 Maths Chapter 13 Question Answers - Statistics

Short Answer Type Questions

Q1. Find the mode of the data:

Ans: Modal class is 20-30.
f₁ = 32, f₀ = 12 and f₂ = 20.
Lower limit l = 20.

      [∵ h = 10]

Use the formula for mode in grouped data:
Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h.
Numerator = 32 - 12 = 20.
Denominator = 2 × 32 - 12 - 20 = 64 - 32 = 32.
Fraction = 20/32 = 0.625.
Mode = 20 + 0.625 × 10 = 20 + 6.25 = 26.25.
Therefore, Mode ≈ 26.25 marks.

Q2. The percentage marks obtained by 100 students in an examination are given below:

Find the median from the above data.

Ans: Total number of observations n = 100, so n/2 = 50.

Here,  

From the cumulative frequencies the median class is 45-50 (first class with cf ≥ 50).
So l = 45, cf = 44 (cumulative frequency before the median class), f = 23 and h = 5.

Use the formula for median in grouped data:
Median = l + [(n/2 - cf)/f] × h.
n/2 - cf = 50 - 44 = 6.
(n/2 - cf)/f = 6/23 ≈ 0.26087.
Multiply by h: 0.26087 × 5 ≈ 1.30435.
Median = 45 + 1.30435 = 46.30435.
Therefore, Median ≈ 46.30%.

Q3. Write a frequency distribution table for the following data:

Ans: To find the frequency in each class, subtract successive 'above' values (Above a - Above b gives number in the class a-b).
30 - 28 = 2
28 - 21 = 7
21 - 15 = 6
15 - 10 = 5
10 - 0 = 10
These are the frequencies for 0-10, 10-20, 20-30, 30-40 and 40-50 respectively.

The required frequency distribution is:

Q4. Find the median of the following data:

Ans:

∵ Median class is 40-60 

So total n = 50 and n/2 = 25.
Thus l = 40, cf = 15 (cumulative frequency before the median class), f = 12 and h = 20.

Use the formula:
Median = l + [(n/2 - cf)/f] × h.
n/2 - cf = 25 - 15 = 10.
(n/2 - cf)/f = 10/12 = 0.83333.
Multiply by h: 0.83333 × 20 = 16.6667.
Median = 40 + 16.6667 = 56.6667.
Therefore, Median ≈ 56.67 marks.