Need Help?

Subscribe to Probability

Subscribe
  • Notes
  • Comments & Questions

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.

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

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

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

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

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

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

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

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

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

$$=1-0.7688=0.2312$$

No comments yet

Join the conversation

Join Notatee Today!