Cholesterol Suppose that the blood cholesterol level of all men aged 20 to 34 follows the Normal distribution with mean $\mu =188$milligrams per deciliter (mg/dl) and standard deviation $\sigma =41\mathrm{mg}/\mathrm{dl}$.

a. Choose an SRS of 100 men from this population. Describe the sampling distribution of ${\mathrm{x}}^{-}·\overline{x}$.

b. Find the probability that $x^{-}\overline{x}$estimates $\mu$within $\pm 3\mathrm{mg}/\mathrm{dl}$. (This is the probability that ${\mathrm{x}}^{-\overline{x}}$takes a value between 185 and$191\mathrm{mg}/\mathrm{dl}$

c. Choose an SRS of 1000 men from this population. Now what is the probability that ${\mathrm{x}}^{-}$ $\overline{x}$ falls within $\pm 3\mathrm{mg}/\mathrm{dl}$ of $\mu$? In what sense is the larger sample "better"?

Given,

Formula used:

${\sigma }_x=\frac{\sigma }{\sqrt{n}}$

Because the population distribution is normal, so is the sampling distribution of the sample mean $\overline{x}$.

${\mu }_{\overline{x}}=\mu =118$

The standard deviation of the sample mean's sampling distribution is

As a result, the sample mean's sampling distribution is Normal, with a mean of 118 and a standard deviation of 4.1.

Given,

$$\begin{gathered} \mu =188 \\ \sigma =41 \\ n=100 \end{gathered}$$

Formula used:

$z=\frac{x-{\mu }_{\overline{z}}}{{\theta }_{\overline{z}}}$

Because the population distribution is normal, so is the sampling distribution of the sample mean $\overline{x}$.

The z-value is

$$\begin{gathered} z=\frac{x-{\mu }_{\overline{z}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{d\sqrt{n}}=\frac{185-188}{41/\sqrt{\overline{100}}}=-0.73 \\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{I}}}=\frac{\overline{x}-\mu }{d\sqrt{n}}=\frac{191-188}{41/\sqrt{\overline{100}}}=-0.7 \end{gathered}$$

The likelihood of associating using the normal probability $P(Z<-0.73)$

The standard normal probability is presented in the row beginning with $-0.7$and the column beginning with $.03$. The standard normal probability $P(Z<0.73)$is presented in the row beginning with 0.7 and the column beginning with $0.3$.

$$\begin{gathered} P(185<\overline{X}<191)=P(-0.73<Z<0.73) \\ =P(Z<0.73)-P(Z<-0.73) \\ =0.7673-0.2327 \\ =0.5346 \\ =53.46\% \end{gathered}$$

Given,

Because the population distribution is normal, so is the sampling distribution of the sample mean $\overline{x}$.

The sample mean of sampling distribution $\overline{x}$has mean $\mu$and standard deviation $\frac{\sigma }{\sqrt{n}}$.

The z-score is

$$\begin{gathered} z=\frac{x-{\mu }_{\overline{\overline{}}}}{{\sigma }_{\overline{\overline{I}}}}=\frac{\overline{\overline{x}}-\mu }{d\sqrt{n}}=\frac{185-188}{41/\sqrt{1000}}=-2.31 \\ z=\frac{x-{\mu }_{\overline{\overline{n}}}}{{\sigma }_{\overline{\overline{x}}}}=\frac{\overline{x}-\mu }{d\sqrt{\sqrt{n}}}=\frac{191-188}{41/\sqrt{1000}}=2.31 \end{gathered}$$

The likelihood of associating using the normal probability The standard normal probability $P(Z<-2.31)$is presented in the row beginning with $-2.3$and the column beginning with $0.01$. $P(Z<2.31)$The standard normal probability is presented in the row beginning with $2.3$and the column beginning with $0.01$.

$$\begin{gathered} P(185<\overline{X}<191)=P(-2.31<Z<2.31) \\ =P(Z<2.31)-P(Z<-2.31) \\ =0.9806-0.0104 \\ =0.9792 \\ =97.92\% \end{gathered}$$

The large sample is "better" because the probability that the sample mean is within three standard deviations of the population mean is greater, and so our estimations (sample means) of the population mean are more accurate.