Question

1. based on data from the national health survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm hg) of 114.8 with a standard deviation of 13.1. (a) if one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120. (b) if 10 women from this population are randomly selected, can you say what the shape of the distribution of the sample mean is? why or why not? (c) if 40 women from this population are randomly selected, can you say what the shape of the distribution of the sample mean is? why or why not? (d) if 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120. (e) if the sample were four women between the ages of 18 to 24, could the central limit theorem be used? explain.

Explanation:

Step1: Standardize for part (a)

We use the z - score formula $z=\frac{x-\mu}{\sigma}$, where $\mu = 114.8$, $\sigma=13.1$ and $x = 120$. So $z=\frac{120 - 114.8}{13.1}=\frac{5.2}{13.1}\approx 0.4$. Then $P(X>120)=1 - P(X\leq120)$. Looking up $P(Z\leq0.4)$ in the standard - normal table, we get $P(Z\leq0.4) = 0.6554$. So $P(X>120)=1 - 0.6554 = 0.3446$.

Step2: Analyze part (b)

We don't know the shape of the population distribution. By the central limit theorem, for the sampling distribution of the sample mean to be approximately normal, we usually need $n\geq30$. Since $n = 10<30$ and we don't know if the population is normal, we cannot say what the shape of the distribution of the sample mean is.

Step3: Analyze part (c)

Since $n = 40\geq30$, by the central limit theorem, the sampling distribution of the sample mean $\bar{X}$ is approximately normal, regardless of the shape of the population distribution.

Step4: Standardize for part (d)

The mean of the sampling distribution of the sample mean is $\mu_{\bar{X}}=\mu = 114.8$ and the standard deviation is $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{13.1}{\sqrt{40}}\approx2.07$. We calculate the z - score: $z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}=\frac{120 - 114.8}{2.07}=\frac{5.2}{2.07}\approx2.51$. Then $P(\bar{X}>120)=1 - P(\bar{X}\leq120)$. Looking up $P(Z\leq2.51)$ in the standard - normal table, we get $P(Z\leq2.51)=0.9940$. So $P(\bar{X}>120)=1 - 0.9940 = 0.0060$.

Step5: Analyze part (e)

The central limit theorem states that for a sample of size $n$ from any population with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample mean is approximately normal when $n\geq30$. Since $n = 4<30$ and we don't know if the population is normal, the central limit theorem cannot be used.

Answer:

(a) $0.3446$
(b) No, because $n = 10<30$ and the population distribution is unknown.
(c) Yes, because $n = 40\geq30$ and by the central limit theorem, the sampling distribution of the sample mean is approximately normal.
(d) $0.0060$
(e) No, because $n = 4<30$ and we don't know if the population is normal.