Question 14 A population of values has a normal distribution with $\mu=198.4$ and $\sigma=74.6$. You intend to draw a random sample of size $n=167$. Find $P_{67}$, which is the score separating the bottom $67 \%$ scores from the top $33 \%$ scores. $P_{67}$ (for single values) $=$ Find $P_{67}$, which is the mean separating the bottom $67 \%$ means from the top $33 \%$ means. $P_{67}$ (for sample means) $=$ Enter your answers as numbers accurate to 1 decimal place. Answers obtained using exact $z$-scores or $z$ scores rounded to 3 decimal places are accepted. Question Help: D Post to forum Submit Question IMG_2842.jpg IMG_2841.jpg IMG 2840.jpg - IMG_0891.jpg

Step 1 ：Given that the population has a normal distribution with mean \(\mu = 198.4\) and standard deviation \(\sigma = 74.6\). We are drawing a random sample of size \(n = 167\).

Step 2 ：We need to find the z-score that corresponds to the 67th percentile in a standard normal distribution. This z-score is approximately \(z_{67} = 0.44\).

Step 3 ：To find the score separating the bottom 67% of individual scores (P67 for single values), we convert this z-score to a value in the population distribution. This gives us \(P_{67\_single} = \mu + z_{67} \times \sigma = 198.4 + 0.44 \times 74.6 \approx 231.2\).

Step 4 ：To find the mean separating the bottom 67% of means (P67 for sample means), we first need to adjust the standard deviation for the sample size. This is called the standard error (SE), and is given by \(SE = \frac{\sigma}{\sqrt{n}} = \frac{74.6}{\sqrt{167}} \approx 5.77\).

Step 5 ：We then repeat the process above using the standard error instead of the standard deviation. This gives us \(P_{67\_sample} = \mu + z_{67} \times SE = 198.4 + 0.44 \times 5.77 \approx 200.9\).

Step 6 ：Final Answer: The score separating the bottom 67% of individual scores (P67 for single values) is approximately \(\boxed{231.2}\). The mean separating the bottom 67% of means (P67 for sample means) is approximately \(\boxed{200.9}\).