Question

The mean of the following distribution is 53. Find the missing frequency p.

Hence, find mode of the distribution.}

Hint:

Double-check the modal class after finding missing frequencies, as it might change depending on the calculated value.

Answer 1

Class intervals: 0–20, 20–40, 40–60, 60–80, 80–100
Frequencies: 12, 15, p, 28, 13
Mean = 53

Step 1: Find mid-values
0–20 → 10
20–40 → 30
40–60 → 50
60–80 → 70
80–100 → 90

Step 2: Use mean formula
Mean = Σ(f·x) / Σf
Σf = 12 + 15 + p + 28 + 13 = 68 + p

Σ(f·x) = 12·10 + 15·30 + p·50 + 28·70 + 13·90
= 120 + 450 + 50p + 1960 + 1170
= 3700 + 50p

Mean = 53, so:
(3700 + 50p) / (68 + p) = 53

Cross multiply:
3700 + 50p = 53(68 + p)
3700 + 50p = 3604 + 53p

Rearranging:
3700 – 3604 = 53p – 50p
96 = 3p
p = 32

Missing frequency is: \[ \boxed{32} \]

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Step 3: Find the mode
Mode is in the class with highest frequency.
Frequencies: 12, 15, 32, 28, 13 → Highest = 32
So modal class = 40–60

Mode formula:
\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right)h \]
l = 40
f₁ = 32
f₀ = 15
f₂ = 28
h = 20

Substitute:
\[ \text{Mode} = 40 + \left( \frac{32 - 15}{2(32) - 15 - 28} \right)20 \] \[ = 40 + \left( \frac{17}{64 - 43} \right)20 \] \[ = 40 + \left( \frac{17}{21} \right)20 \] \[ = 40 + \frac{340}{21} \] \[ = 40 + 16.19 = 56.19 \]

Final Answers:
Missing frequency p = \[ \boxed{32} \]
Mode ≈ \[ \boxed{56.19} \]