91影视

Given in the question that

Population mean $\left(渭\right)=188$

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

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

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

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

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

$\stackrel{炉}{x}~N\left(渭,\frac{蟽}{\sqrt{n}}\right)$

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

$\stackrel{炉}{x}~N(188,4.1)$

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

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

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

$P(185鈮�\overline{x}<191)=P\left(\frac{{\displaystyle \frac{185-渭}{蟽}}}{\sqrt{n}}<\frac{{\displaystyle \frac{\overline{x}-渭}{蟽}}}{\sqrt{n}}<\frac{{\displaystyle \frac{191-渭}{蟽}}}{\sqrt{n}}\right)$

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

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

$=$ $0.5346$

Thus,the required probability is$0.5346$

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

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

$P\left(185鈮�\overline{x}<191\right)=P\left(\frac{{\displaystyle \frac{185-渭}{蟽}}}{\sqrt{n}}<\frac{{\displaystyle \frac{\stackrel{炉}{x}-渭}{蟽}}}{\sqrt{n}}<\frac{{\displaystyle \frac{191-渭}{蟽}}}{\sqrt{n}}\right)$

$=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 $0.9792$