Question

The mean and s.D. Based on 60 observations were calculated as 276 mm and 22 mm respectively. But on subsequent scrutiny it was found that through oversight one observation was recorded wrongly at 327 mm through its correct value was 372 mm. Find the corrected values of the mean and s.D.

Answer 1

Correct mean is 276.75 mm and correct Standard deviation is 24.37 mm.

• Given:

Number of observations = 60

Incorrect mean = 276 mm

Incorrect Standard deviation = 22 mm

• We know that ,

Mean = $\frac{1}{n}\sum_{i=1}^{n} x_i$

Mean = $\frac{Incorrect \ Sum \ of \ all \ the \ numbers}{n}$

276 = $\frac{Incorrect \ Sum \ of \ all \ the \ numbers}{60}$

Incorrect Sum of all numbers = 16560

∴ Incorrect sum of observations = 16560

• Finding correct sum of observations,

327 was taken in place of 372.

• Correct sum of observations = incorrect sum - 327 + 372 = 16605
• Correct mean = 16605 / 60 = 276.75 mm

Now, Incorrect SD = 22 mm

SD = $\frac{1}{n}\sqrt{n (Incorrect \sum x_i^2) - (Incorrect \sum x_i)^2}$

22 = $\frac{1}{60}\sqrt{60 (Incorrect \sum x_i^2) - (16560)^2}$

1320= $\sqrt{60 (Incorrect \sum x_i^2) - (16560)^2}$

1742400 = ${60 (Incorrect \sum x_i^2) - 27423360}$

1742400 + 27423360 = $60 (Incorrect \sum x_i^2)$

275976000 = $60 (Incorrect \sum x_i^2)$

275976000 / 60 = $(Incorrect \sum x_i^2)$

$(Incorrect \sum x_i^2)$ = 4599600

• Since 327 is removed and 372 is added,

$(correct \sum x_i^2)$ = 4599600 - 327² + 372²

$(correct \sum x_i^2)$ = 4631055

• Now,

Correct SD = $\frac{1}{n}\sqrt{n (correct \sum x_i^2) - (correct \sum x_i)^2}$

Correct SD = $\frac{1}{60}\sqrt{60 (4631055) - (16605)^2}$

Correct SD = $\frac{1}{60}\sqrt{277863300 - ‭275726025‬}$

Correct SD = 24.37 mm