Question

1. 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. How is the standard deviation of the sample mean changed when the sample size is increased from n=4 to n=52? Round all intermediate calculations to four decimal places (e.g. 12.3456) and round the final answer to three decimal places (e.g. 98.768).

The standard deviation is ( reduced, increased ) by _____ psi.

2. Suppose that X has a discrete uniform distribution

f(x) = {1/3, x=1,2,3
{ 0, otherwise
A random sample of n= 38 is selected from this population. Find the probability that the sample mean is greater than 2.1 but less than 2.4. Express the final answer to four decimal places (e.g. 0.9876).

The probability is _______

Answer 1

1. $X\sim N(\mu, \sigma^2/n)$

Given $\mu=75.5, \sigma=3.5,n_1=2,n_2=52$

$$\sigma/\sqrt{n_2}-\sigma/\sqrt{n_2}=3.5/\sqrt{52}-3.5/\sqrt{4}\approx-1.2646$$

The standard deviation is reduced by $1.2646$ psi. 

2.

$$f(x) = \begin{cases} 1/3, & x=1,2,3 \\ 0, &\text{otherwise } \end{cases}$$

Given $n=38, a=1, b=3.$

Sample mean

$$\mu_{\bar{X}}={b+a \over 2}={3+1 \over 2}=2$$

Standard deviation

$$\sigma_{\bar{X}}=\sqrt{{(b-a+1)^2-1 \over 12}}=\sqrt{{(3-1+1)^2-1 \over 12}}=\sqrt{{2 \over 3}}\approx0.8165$$

$$P(2.1<\bar{X}<2.4)=P(\dfrac{2.4-\mu_{\bar{X}}}{\sigma_{\bar{X}}/\sqrt{n}}<Z<\dfrac{2.1-\mu_{\bar{X}}}{\sigma_{\bar{X}}/\sqrt{n}})=$$

$$=P(\dfrac{2.4-2}{0.8165/\sqrt{38}}<Z<\dfrac{2.1-2}{0.8165/\sqrt{38}})\approx$$

$$\approx P(Z<3.0199)-P(Z\leq0.7550\approx$$

$$\approx0.99874-0.77488\approx0.2239$$

The probability is $0.2239$