Question

Larger sample Suppose that the blood cholesterol level of all men aged $20-34$follows the Normal distribution with mean $\mu =188$milligrams per deciliter (mg/dl) and standard deviation $\sigma =41mg/dl$

(a) Choose an SRS of $100$men from this population What is the sampling distribution of $\overline{x}$?

(b) Find the probability that $\overline{x}$estimates $\mu$within $\pm 3mg/dl$. (This is the probability that$\overline{x}$takes a value between $185and191mg/dl$.) Show your work.

(c) Choose an SRS of $1000$men from this population. Now what is the probability that $\overline{x}$falls within$\pm 3mg/dl$of$\mu$? show your wrok.in what sense is the large sample "better".

Short answer

(a) The sampling distribution is normally distributed with the mean $=100$and standard deviation$=4.1$

(b) The probability is $0.5346$

(c) The probability is$0.9792$

Step-by-step solution

Part (a) Step-1 Given Information

Given in the question that

Population mean $\left(\mu \right)=188$

Population standard deviation $\left(\sigma \right)=41$

Sample size $\left(n\right)=100$

we have to find out that What is the sampling distribution of $\overline{x}$

Part (a) Step-2 Explanation

The sample distribution of $\overline{x}$is written as:

role="math" localid="1649195702118" $\overline{x}~N({\mu }_{\overline{X}},{\sigma }_{\overline{x}})$

$\overline{x}~N\left(\mu ,\frac{\sigma }{\sqrt{n}}\right)$

$\overline{x}~N\left(188,\frac{41}{\sqrt{100}}\right)$

$\overline{x}~N(188,4.1)$

The sampling distribution is normally distributed with the mean $100$ and standard deviation$4.1$

Part (b) Step-1: Given Information 

Given in the question that the probability that$\overline{x}$takes a value between $185and191mg/dl$we have to find the probability that $\overline{x}$estimates $\mu$within $\pm 3mg/dl$.

Part (b) Step-2 Explanation

The probability that mean is within $\pm 3mg/dl$is calculated as follows:

$P(185\le \overline{x}<191)=P\left(\frac{{\displaystyle \frac{185-\mu }{\sigma }}}{\sqrt{n}}<\frac{{\displaystyle \frac{\overline{x}-\mu }{\sigma }}}{\sqrt{n}}<\frac{{\displaystyle \frac{191-\mu }{\sigma }}}{\sqrt{n}}\right)$

$=P\left(\frac{{\displaystyle \frac{185-191}{41}}}{\sqrt{100}}<Z<\frac{{\displaystyle \frac{191-188}{41}}}{\sqrt{100}}\right)$

$=P(-0.73<Z<0.73)$

$=$ $0.5346$

Thus,the required probability is$0.5346$

Part (c) Step-1 Given Information 

Given in the question that choose an SRS of $1000$men from this population. we have to find the probability that $\overline{x}$falls within $\pm 3mg/dl$of$\mu$.

Part (c) Step-2: Explanation 

The probability that mean is within $\pm 3mg/dl$is calculated as follows:

$P\left(185\le \overline{x}<191\right)=P\left(\frac{{\displaystyle \frac{185-\mu }{\sigma }}}{\sqrt{n}}<\frac{{\displaystyle \frac{\overline{x}-\mu }{\sigma }}}{\sqrt{n}}<\frac{{\displaystyle \frac{191-\mu }{\sigma }}}{\sqrt{n}}\right)$

$=P\left(\frac{{\displaystyle \frac{185-191}{41}}}{\sqrt{1000}}<Z<\frac{{\displaystyle \frac{191-188}{41}}}{\sqrt{1000}}\right)$$=P\left(\frac{{\displaystyle \frac{185-188}{41}}}{\sqrt{1000}}<Z<\frac{{\displaystyle \frac{191-188}{41}}}{\sqrt{1000}}\right)$

$=P(-2.31<Z<2.31)$

$=$$0.9792$$=0.9792$

Thus the required probability is $0.9792$