Questions: A simple random sample of size n=81 is obtained from a population that is skewed right with μ=74 and σ=27. (a) Describe the sampling distribution of x̄. (b) What is P(x̄>78.2) ? (c) What is P(x ≤ 66.95) ? (d) What is P(69.5<x̄<79.25) ?

Solution Steps

The sampling distribution of the sample mean $\bar{x}$ is approximately normal due to the Central Limit Theorem. It has the following parameters:

• Mean: $\mu = 74$
• Standard Deviation: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{27}{\sqrt{81}} = 3.0$

Thus, the sampling distribution can be expressed as: $$\bar{x} \sim N(74, 3.0)$$

To find $P(\bar{x} > 78.2)$, we first convert $78.2$ to a Z-score: $$Z = \frac{78.2 - \mu}{\sigma_{\bar{x}}} = \frac{78.2 - 74}{3.0} = 1.4$$ Using the standard normal distribution, we find: $$P(\bar{x} > 78.2) = 1 - P(Z \leq 1.4) = 1 - 0.9192 = 0.0808$$

Next, we calculate $P(\bar{x} \leq 66.95)$ by converting $66.95$ to a Z-score: $$Z = \frac{66.95 - \mu}{\sigma_{\bar{x}}} = \frac{66.95 - 74}{3.0} = -2.35$$ Thus, we have: $$P(\bar{x} \leq 66.95) = P(Z \leq -2.35) = 0.0094$$

Final Answer