Question

A random sample of size n=68 is taken from a finite population of size N=644 with mean y = 239 and variance o?. 325. [You may find it useful to reference the table.) 8-1. Is it necessary to apply the finite population correction factor? Yes No 0-2. Calculate the expected value and the standard error of the sample mean (Round "expected value to a whole number and standard error to 4 decimal places.) Answer is complete but not entirely correct. Expected value Standard error 239 2.1851 b. What is the probability that the sample mean is less than 228? (Round " places.) value to 2 decimal places, and final answer to 4 decimal Answer is complete and correct. 0.0000 Probability c. What is the probability that the sample mean lies between 236 and 2517 (Use rounded standard deviation Round " value to 2 decimal places and final answer to 4 decimal places)

Answer 1

A sample of size n=68 is taken from a population of size N=644

a-1) We use the finite population correction factor when the sample size is 5% or more than the population

The sampling fraction in this case is

$\frac{n}{N}=\frac{68}{644}=0.1056$

This sampling fraction is greater than 0.05. Hence we need to use the finite population correction factor

ans: Yes

a-2) Let $\bar{X}$ be a sample of size n=68 taken from a population of size N=644, having a mean $\mu=239$ and variance $\sigma^2=325$

We can say that the expected value of $\bar{X}$ is

$E(\bar{X})=239$

the standard deviation of $\bar{X}$ (ir the standard error of mean ) is

$\sigma_{\bar{X}}=\sqrt{\frac{\sigma^2}{n}\times \frac{N-n}{N-1}}=\sqrt{\frac{325}{68}\times \frac{644-68}{644-1}}=2.0692$

ans:

• Expected value: 239
• Standard error: 2.0692

b) Since the sample size n=68 is greater than 30, using the central limit theorem, we can say that $\bar{X}$ has a normal distribution with mean $\mu_{\bar{X}}=239$ and standard deviation (or is called the standard error of mean) $\sigma_{\bar{X}}=2.0692$

The probability that the sample mean is less than 228 is

$\begin{align*} P(\bar{X}<228)&=P\left(\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}<\frac{228-\mu_{\bar{X}}}{\sigma_{\bar{X}}} \right )\quad\text{get the z score of 228}\\ &=P\left(Z<\frac{228-239}{2.0692} \right )\\ &=P(Z<-5.32)\\ &=P(Z>5.32)\\ &=1-P(Z<5.32)\\ &=1-1.000\quad\text{using the standard normal table for z=5.32}\\ &=0.0000 \end{align*}$

ans: The probability that the sample mean is less than 228 is 0.0000

c) The probability that the sample mean is between 236 and 251 is

$\begin{align*} P(236<\bar{X}<251)&=P\left(\frac{236-\mu_{\bar{X}}}{\sigma_{\bar{X}}}<\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}<\frac{251-\mu_{\bar{X}}}{\sigma_{\bar{X}}} \right )\quad\text{get the z score of 236,251}\\ &=P\left(\frac{236-239}{2.0692}<Z<\frac{251-239}{2.0692} \right )\\ &=P(-1.45<Z< 5.80)\\ &=P(Z<5.80)-P(Z<-1.45)\\ &=P(Z<5.80)-P(Z>1.45)\\ &=P(Z<5.80)-(1-P(Z<1.45))\\ &=1.000-(1-0.9265)\quad\text{using the standard normal table for z=5.80,1.45}\\ &=0.9265 \end{align*}$

ans: probability: 9265