Question

question #1
for a population, the mean is $360 with a standard deviation of $87. determine the probability of a sample of 21 having a mean between $400 and $450.
.0561
.0176
.2239
.1724

question #2
for a population, the mean is $1,087 with a standard deviation of $305. determine the probability of a sample of 45 having a mean of $895 or lower.
0.0001
0.4685
0.5315
0.9999

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$.
For Question #1: $\sigma = 87$, $n = 21$, so $\sigma_{\bar{x}}=\frac{87}{\sqrt{21}}\approx19.07$.
For Question #2: $\sigma = 305$, $n = 45$, so $\sigma_{\bar{x}}=\frac{305}{\sqrt{45}}\approx45.47$.

Step2: Calculate the z - scores

The formula for the z - score is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$.
For Question #1:
When $\bar{x}=400$, $z_1=\frac{400 - 360}{19.07}\approx2.09$.
When $\bar{x}=450$, $z_2=\frac{450 - 360}{19.07}\approx4.72$.
Then $P(2.09Since $\Phi(4.72)\approx1$ and $\Phi(2.09) = 0.9817$, $P(2.09For Question #2:
When $\bar{x}=895$, $z=\frac{895 - 1087}{45.47}=\frac{- 192}{45.47}\approx - 4.22$.
Then $P(Z\leq - 4.22)\approx0.0001$.

Answer:

Question #1: 0.0176
Question #2: 0.0001