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A synthetic fiber used in manufacturing carpet has
tensile strength that is normally distributed with mean 75.5 psi
and standard deviation 3.5 psi. Find the probability that a ran-
dom sample of $n=6$ fiber specimens will have sample mean
tensile strength that exceeds 75.75 psi.

tensile strength $\rightarrow$ normal distribution

$\mu=75.5 \mathrm{Psi}$

$\sigma=3.5 \mathrm{Psi}$

sample of $n=6$

$P(\overline x>75.75)$

$\mu_{x}=\mu=75.5$

$\sigma_{\overline x}=\frac{\sigma}{\sqrt{n}}=\frac{3.5}{\sqrt{6}}$

$z=\frac{\overline{x}-\mu_{\overline{x}}}{\sigma_{\overline{x}}}$

$=\frac{75.75-75.5}{3.5 / \sqrt{6}}=0.175$

$P(z>0.175)=1 - P(z<0.175)$

$\phi(0.175)$

$=1-0.56945=0.43055$

Suppose that $X$ has a discrete uniform distribution

$f(x)=\left\{\begin{array}{cc}{1 / 3,} & {x=1,2,3} \\ {0,} & {\text { otherwise }}\end{array}\right.$

A random sample of $n=36$ is selected from this population.
Find the probability that the sample mean is greater than 2.1
but less than $2.5,$ assuming that the sample mean would be
measured to the nearest tenth.

$\mu_{x}=\frac{a+b}{2}=\frac{3+1}{2}=2$

$\sigma_{x}=\sqrt{\frac{(b-a+1)^{2}-1}{12}}=\sqrt{\frac{8}{12}}=\sqrt{\frac{2}{3}}$

$\mu_{\overline{x}}=\mu = 2$

$\sigma_{\overline{x}}=\frac{\sigma_{x}}{\sqrt{n}}$

$=\frac{\sqrt{2 / 3}}{\sqrt{36}}=\frac{\sqrt{2 / 3}}{6}$

$z=\frac{\overline{x}-\mu}{\sigma / \sqrt{n}}$

$P\left(\frac{2.1-2}{\frac{\sqrt {2/3}}{6}}<z<\frac{2.5-2}{\frac{\sqrt{2/3}}{6}}\right)$

$P(0.7348<z<3.6742)$

$=P(z<3.6742)-P(z<0.7348)$

$=1-0.7688=0.2312$