The weekly incomes of the workers of a large city have a skewed distribution with mean Rs. 5,000 and standard deviation Rs. 1,200. Random samples of 144 workers each are drawn. By the central limit theorem, the sampling distribution of the sample mean is approximately
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → normal, with mean Rs. 5,000 and standard deviation Rs. 100
By the central limit theorem, for a large sample (n = 144, well above 30) the sampling distribution of the sample mean is approximately normal, whatever the shape of the population.
Its mean is the population mean, Rs. 5,000.
Its standard deviation is the standard error of the mean:
\( \displaystyle \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1200}{\sqrt{144}} \)
\( \displaystyle = \frac{1200}{12} = 100 \)
Why the other options are incorrect
Option (2) normal, with mean Rs. 5,000 and standard deviation Rs. 1,200 — Incorrect. Rs. 1,200 is the spread of single incomes; sample means vary much less, by \(\sigma/\sqrt{n} =\) Rs. 100.
Option (3) skewed like the population, with mean Rs. 5,000 and standard deviation Rs. 100 — Incorrect. For a large sample the sampling distribution of the mean is approximately normal even when the population is skewed.
Option (4) normal, with mean Rs. 5,000 and standard deviation Rs. 8.33 — Incorrect. 8.33 is 1200 ÷ 144, dividing by n instead of \(\sqrt{n} = 12\).