Question

The Lakeland Post polled 1,231 adults in the city to determine whether they wash their hair before their body while in the shower. Of the respondents, 63% said they washed their hair first. Suppose 51% of all adults actually wash their hair first. What are the mean and standard deviation of the sampling distribution? A. The mean is 0.51 and the standard deviation is 0.0142. B. The mean is 0.51 and the standard deviation is 0.0138. C. The mean is 0.63 and the standard deviation is 0.0226. D. The mean is 0.63 and the standard deviation is 0.0138. E. The mean is 0.63 and the standard deviation is 0.0142.

Answer 1

Answer:

A. The mean is 0.51 and the standard deviation is 0.0142.

Step-by-step explanation:

When we know the mean of the population (0.51), we don't need to estimate the mean of the sampling distribution from the mean of a sample (0.63).

Then, the mean of the sampling distribution is equal to the mean of the population (0.51).

The standard deviation is calculated as:

$\sigma_p=\sqrt{\frac{\pi(1-\pi)}{N} \frac{x}{y} }= \sqrt{\frac{0.51(1-0.51)}{1231} }=0.0142$