GATE 2023 IN Q.53: solution and answer

GATE 2023 · INQ.53 · 2 marks · NAT

Five measurements are made using a weighing machine, and the readings are 80 kg, 79 kg, 81 kg, 79 kg and 81 kg. The sample standard deviation of the measurement is ________ kg (rounded off to two decimal places).

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Answer and solution

ANSWER

The sample standard deviation is $s = 1.00 kg$ .

Official answer: 0.98 to 1.02Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Sample mean: $\overset{x}{ˉ} = \frac{80 + 79 + 81 + 79 + 81}{5} = \frac{400}{5} = 80 kg$ .
Deviations from the mean, $x_{i} - \overset{x}{ˉ}$ : $0, - 1, + 1, - 1, + 1$ .
Sum of squared deviations: $\sum (x_{i} - \overset{x}{ˉ})^{2} = 0 + 1 + 1 + 1 + 1 = 4 kg^{2}$ .
The question asks for the sample standard deviation, so divide by $n - 1$ (Bessel's correction), not by $n$ : $s = \frac{\sum ( x _{i} - x ˉ ) ^{2}}{n - 1} = \frac{4}{5 - 1} = 1 = 1 kg$
Rounded off to two decimal places, $s = 1.00 kg$ .

KEY IDEA

The sample standard deviation divides the sum of squared deviations by $n - 1$ (Bessel's correction), not by $n$ : $s = \sum (x_{i} - \overset{x}{ˉ})^{2} / (n - 1)$ . Here the mean is 80 kg, the squared deviations add up to 4, and $s = 4/4 = 1$ .

WATCH OUT FOR

Dividing by $n = 5$ gives the population standard deviation, $4/5 \approx 0.89 kg$ . That falls outside the key's range of 0.98 to 1.02. The word "sample" in the question means you divide by $n - 1 = 4$ .

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GATE 2023 IN Q.53: solution and answer

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