The mean and standard deviation of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that one student's marks 40 was wrongly read as 50. The correct mean and standard deviation, respectively are

Question

Smart Achievers · Accepted Answer

(2)

$Corrected \sum x = 40 \times 200 - 50 + 40 = 7990$

$∴$ $Corrected \bar{x} = \frac{7990}{200} = 39.95$

$Incorrect \sum x^{2} = n [σ^{2} + {\bar{x}}^{2}] = 200 [15^{2} + 40^{2}] = 365000$

$Correct \sum x^{2} = 365000 - 2500 + 1600 = 364100$

$∴$ $Corrected σ = \sqrt{\frac{364100}{200} - {(39.95)}^{2}} = \sqrt{1820.5 - 1596} = \sqrt{224.5} = 14.98$

Solution

Q.
The mean and standard deviation of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that one student's marks 40 was wrongly read as 50. The correct mean and standard deviation, respectively are

1 14.98, 39.95

2 39.95, 14.98

3 39.95, 224.5

4 39.00, 14.00

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