Answer to Question #190102 in Statistics and Probability for Ahmad

Question

Household size in the United States has a mean of 2.6 people and standard deviation of 1.4 people. It should be clear that this distribution is skewed right as the smallest possible value is a household of 1 person but the largest households can be very large indeed.

(a) What is the probability that a randomly chosen household has more than 3 people?

A normal approximation should not be used here, because the distribution of household sizes would be considerably skewed to the right. We do not have enough information to solve this problem.

(b) What is the probability that the mean size of a random sample of 10 households is more than 3?

By anyone’s standards, 10 is a small sample size. The Central Limit Theorem does not guarantee sample mean coming from a skewed population to be approximately normal unless the sample size is large.

(c) What is the probability that the mean size of a random sample of 100 households is more than 3?

Answer 1

(a) Already answered

A normal approximation should not be used here, because the distribution of household sizes would be considerably skewed to the right. We do not have enough information to solve this problem.

(b) Already answered

A normal approximation should not be used here, because the distribution of household sizes would be considerably skewed to the right. We do not have enough information to solve this problem.

(c) Now we may invoke the Central Limit Theorem: even though the distribution of household size X is skewed, the distribution of sample mean household size $\bar X$ is approximately normal for a large sample size such as 100. Its mean is the same as the population mean, 2.6, and its standard deviation is the population standard deviation divided by the square root of the sample size:

$$\dfrac{\sigma}{\sqrt n}=\dfrac{1.4}{\sqrt {100}}=0.14$$

To find $P(\bar X>3)$

we standardize 3 to into a z-score by subtracting the mean and dividing the result by the standard deviation (of the sample mean). Then we can find the probability using the standard normal calculator or table.

$$P(\bar X>3)=P(Z>\dfrac{3-2.6}{\frac{1.4}{\sqrt {100}}})=P(Z>2.86)=0.0021$$

So, households of more than 3 people are, of course, quite common, but it would be extremely unusual for the mean size of a sample of 100 households to be more than 3.