Question

1. multiple choice 4 points find the indicated probability. assume that the salaries of elementary school teachers in the united states are normally distributed with a mean of $32,000 and a standard deviation of $3000. if 100 teachers are randomly selected, find the probability that their mean salary is greater than $32,500. 0.1312 0.0475 0.9525 0.3312 25. multiple choice 4 points find the indicated probability. the average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22 pounds (source: american dietetic association). if a sample of 50 individuals is randomly selected, find the probability that the mean of the sample will be greater than 200 pounds. 0.9015 0.7613 0.0985 0.8815

Explanation:

Step1: Identify the distribution and parameters for question 24

The salaries of elementary - school teachers are normally distributed with mean $\mu = 32000$ and standard deviation $\sigma = 3000$, and the sample size $n = 100$. The sampling distribution of the sample mean $\bar{X}$ has mean $\mu_{\bar{X}}=\mu = 32000$ and standard deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{3000}{\sqrt{100}} = 300$.

Step2: Calculate the z - score

We want to find $P(\bar{X}>32500)$. The z - score is calculated as $z=\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}=\frac{32500 - 32000}{300}=\frac{500}{300}\approx1.67$.

Step3: Find the probability using the standard normal table

$P(\bar{X}>32500)=P(Z > 1.67)=1 - P(Z\leqslant1.67)$. From the standard - normal table, $P(Z\leqslant1.67)=0.9525$, so $P(Z > 1.67)=1 - 0.9525 = 0.0475$.

Step4: Identify the distribution and parameters for question 25

The population mean $\mu = 196$, the population standard deviation $\sigma = 22$, and the sample size $n = 50$. The sampling distribution of the sample mean $\bar{X}$ has mean $\mu_{\bar{X}}=\mu = 196$ and standard deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{22}{\sqrt{50}}\approx3.11$.

Step5: Calculate the z - score

We want to find $P(\bar{X}>200)$. The z - score is $z=\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}=\frac{200 - 196}{3.11}\approx1.29$.

Step6: Find the probability using the standard normal table

$P(\bar{X}>200)=P(Z > 1.29)=1 - P(Z\leqslant1.29)$. From the standard - normal table, $P(Z\leqslant1.29)=0.9015$, so $P(Z > 1.29)=1 - 0.9015 = 0.0985$.

Answer:

1. 0.0475
2. 0.0985