Median will change if observations are changes but standard deviation is unaffected by origin. So, here median will go up by 4 and S.D. will remain same.

Combined Mean = X = (n₁x₁ + n₂x₂)/(n₁ + n₂) = (250 × 20 + 320 × 12)/650 = 13.6
Let d₁ = x₁ – X = 20-13.6 = 6.4, d₂ = x₂ – X = 12-13.6 = -1.6
Variance = [250(6.4 + 40.96) + 320(13.6 + 2.56)]/650 = 26.17.

\(\frac{1}{5}\) (Σ x_i) = 3 = 1 + 3 + 5 + x₄ + x₅
⇒ x₄ + x₅ = 15 ———— (i)
Now, \(\frac{1}{5}\) Σ (x_i)² ? \([\frac{1}{5}\) Σ x_i]² = 4
⇒ (x₄)² + (x₅)² = 30 —————-(ii)
Solving equation i and ii, x₄ = 2 and x₅ = 4.

Mean = (117 + 156 + 206 + 198 + 223)/5 = 180

Mean deviation = \(\frac{1}{n}\)[Σ_{i = 5} |x_I – mean|] = 1/5 x [(-63) + (-24) + 26 + 18 + 43] = 1/5 x [0] = 0.

S.D. = √variance
? (0.012)² = Variance
? Variance = 0.000144

(σ_x)² = h²(σ_u)², if u = (x – h)/a
Now, h = (1/5). ⇒ Variance, (σ_u)² = 9 × 25 = 225

Change in origin does not affect standard deviation, whereas standard deviation is affected by scale.

The standard deviation is independent of change of origin. So, standard deviation for new set will remain same.

Coefficient of Variance = (Standard Deviation/Mean) × 100
⇒ Coefficient of Variance = (820/50) × 100 = 16.4.

From given data,

Now, Mean = \(\frac{1}{n}\) Σ f_ix_i = 1080/60 = 18
⇒ Mean deviation = \(\frac{1}{n}\) Σ f_i|x_i – 18| = 462/60 = 7.7.