Question

For a group of 100 candidates, the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on, it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation

Answer 1

Number of candidates = 100

n = 100

Mean `(barx)` = 60

standard deviation (σ) = 15

Mean `(barx) =  (sumx)/"n"`

⇒ 60 = `(sumx)/100`

`sumx` = 6000

Correct total = 6000 + (45 – 40) + (72 – 27)

= 6000 + 5 + 45

= 6050

Correct mean `(barx)` = `6050/100` = 60.5

Given standard deviation = 15

Standard deviation (σ) = `sqrt((sumx^2)/"n" - ((sumx)/"n")^2`

15 = `sqrt((sumx^2)/100 - (60)^2`

Squaring on both sides we get,

225 = `(sumx^2)/100 - (60)^2`

⇒ 225 = `(sumx^2)/100 - 3600`

∴ `(sumx^2)/100` = 225 + 3600 = 3825

`sumx^2` = 3825 × 100

`sumx^2` = 382500

Correct value of `sumx^2` = 382500 + 45² + 72² – 40² – 27²

= 382500 + 2025 + 5184 – 1600 – 729

= 389709 – 2329

= 387380

Correct standard deviation (σ) = `sqrt((sumx^2)/"n" - ((sumx)/"n")^2`

= `sqrt(387380/100 - (60.5)^2`

= `sqrt(3873.8 - 3660.25)`

= `sqrt(213.55)`

= 14.613

⇒ 14.61

Correct mean = 60.5

Correct standard deviation (σ) = 14.61