Suppose a random sample of size $n=36$ is taken from a population with unknown mean and known standard deviation $\sigma =12$. The sample mean is $\overline{x}=50$. Calculate the standard error of the mean (rounded to 4 decimal places). May normality be assumed for the sampling distribution of the mean?

• A. Standard error = $12.0000$; Yes, normality may be assumed because the population standard deviation is known.
• B. Standard error = $0.3333$; No, normality may not be assumed because the population distribution is unknown.
• C. Standard error = $6.0000$; No, normality may not be assumed because the sample size is small.
• D. Standard error = $2.0000$; Yes, normality may be assumed because the sample size is large.

Answer: (D)

Verified step by step guidance

1. Identify the given information: sample size \(n = 36\), population standard deviation \(\sigma = 12\), and sample mean \(\bar{x} = 50\) (though the sample mean is not needed to calculate the standard error).
2. Recall the formula for the standard error of the mean (SEM) when the population standard deviation is known: \(\text{SEM} = \frac{\sigma}{\sqrt{n}}\)
3. Substitute the known values into the formula: \(\text{SEM} = \frac{12}{\sqrt{36}}\)
4. Simplify the denominator by calculating the square root of the sample size: \(\sqrt{36} = 6\)
5. Divide the population standard deviation by this value to find the standard error: \(\text{SEM} = \frac{12}{6}\)
6. Regarding normality: Since the sample size is large (usually \(n \geq 30\) is considered large), by the Central Limit Theorem, the sampling distribution of the sample mean can be assumed to be approximately normal even if the population distribution is unknown.