1. **Problem Statement:** We have a population mean $\mu = 17$ minutes and population standard deviation $\sigma = 12$ minutes. A sample of size $n = 50$ is taken. We assume the sample mean $\overline{x}$ is normally distributed. (A) Find $P(\overline{x} \geq 15)$. (B) Find the value $x$ such that $P(\overline{x} < x) = 0.67$. (C) Discuss if the answers hold if the population distribution is skewed right. 2. **Formulas and Important Rules:** - The sampling distribution of the sample mean $\overline{x}$ has mean $\mu_{\overline{x}} = \mu$ and standard deviation (standard error) $\sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}$. - For normal distribution, probabilities can be found using the standard normal variable $Z = \frac{\overline{x} - \mu}{\sigma_{\overline{x}}}$. - If the population is normal, the sample mean distribution is normal regardless of $n$. - If population is not normal but $n$ is large (usually $n \geq 30$), by Central Limit Theorem, sample mean distribution is approximately normal. 3. **Calculate standard error:** $$\sigma_{\overline{x}} = \frac{12}{\sqrt{50}} = \frac{12}{7.071} \approx 1.697$$ 4. **Part (A): Find $P(\overline{x} \geq 15)$** - Calculate $Z$ for $\overline{x} = 15$: $$Z = \frac{15 - 17}{1.697} = \frac{-2}{1.697} \approx -1.179$$ - Probability $P(\overline{x} \geq 15) = P(Z \geq -1.179)$. - Using symmetry of normal distribution: $$P(Z \geq -1.179) = 1 - P(Z < -1.179) = 1 - \Phi(-1.179)$$ - But $\Phi(-1.179) = 1 - \Phi(1.179)$, so: $$P(Z \geq -1.179) = 1 - (1 - \Phi(1.179)) = \Phi(1.179)$$ - From standard normal tables or calculator, $\Phi(1.179) \approx 0.880$. 5. **Part (B): Find $x$ such that $P(\overline{x} < x) = 0.67$** - Find $Z$ such that $P(Z < z) = 0.67$. - From standard normal tables, $z \approx 0.44$. - Convert back to $x$: $$x = \mu + z \cdot \sigma_{\overline{x}} = 17 + 0.44 \times 1.697 \approx 17 + 0.747 = 17.747$$ 6. **Part (C): Should we trust the answers if population is skewed right?** - Since $n=50$ is large, by Central Limit Theorem, the sampling distribution of the mean is approximately normal even if population is skewed. - Therefore, the answers in (A) and (B) are still approximately correct. **Final answers:** - (A) $P(\overline{x} \geq 15) \approx 0.88$ - (B) The value is approximately $17.75$ minutes. - (C) We should stick with the answers because the sample size is large enough for the Central Limit Theorem to apply.