Q.53 - The Practice of Statistics for AP

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 $\sigma =41 mg/dl$

(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 $\pm 3 mg/dl$. (This is the probability that$\overline{x}$ takes a value between $185 and 191 mg/dl$ .) Show your work.

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

Step-by-Step Solution

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 $0.9792$

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:

$\overline{x}~N ({\mu }_{\overline{X}},{\sigma }_{\overline{x}})$

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

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

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

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 $185 and 191 mg/dl$ we have to find the probability that $\overline{x}$ estimates $\mu$ within $\pm 3 mg/dl$. 

Part (b) Step-2 Explanation

The probability that mean is within $\pm 3 mg/dl$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 $\pm 3 mg/dl$of$\mu$.

Part (c) Step-2: Explanation

The probability that mean is within $\pm 3 mg/dl$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 $0.9792$