core: $66.6 / 100 \quad 10 / 17$ answered
Question 5
A population of values has a normal distribution with $\mu=96$ and $\sigma=79.4$.
a. Find the probability that a single randomly selected value is between 103 and 105.8 . Round your answer to four decimal places.
\[
P(103< X< 105.8)=
\]
b. Find the probability that a randomly selected sample of size $n=128$ has a mean between 103 and 105.8. Round your answer to four decimal places.
\[
P(103< M< 105.8)=
\]

Step 1 ：We are given a population of values that has a normal distribution with a mean (\(\mu\)) of 96 and a standard deviation (\(\sigma\)) of 79.4.

Step 2 ：We are asked to find the probability that a single randomly selected value is between 103 and 105.8. To do this, we first need to convert these values to z-scores using the formula \(z = \frac{X - \mu}{\sigma}\).

Step 3 ：Calculating the z-scores for 103 and 105.8, we get approximately 0.088 and 0.123 respectively.

Step 4 ：We then look up these z-scores in a standard normal distribution table to find the corresponding probabilities, which are approximately 0.535 and 0.549 respectively.

Step 5 ：The probability that a single randomly selected value is between 103 and 105.8 is the difference between these two probabilities, which is approximately \(\boxed{0.0140}\).

Step 6 ：We are also asked to find the probability that a randomly selected sample of size 128 has a mean between 103 and 105.8. The distribution of the sample mean is also normally distributed, but with a standard error (SE) instead of the population standard deviation. The SE is calculated as \(\sigma / \sqrt{n}\), where n is the sample size.

Step 7 ：Calculating the SE for a sample size of 128, we get approximately 7.018.

Step 8 ：We then calculate the z-scores for 103 and 105.8 using the SE, which are approximately 0.997 and 1.396 respectively.

Step 9 ：Looking up these z-scores in a standard normal distribution table, we find the corresponding probabilities to be approximately 0.841 and 0.919 respectively.

Step 10 ：The probability that a randomly selected sample of size 128 has a mean between 103 and 105.8 is the difference between these two probabilities, which is approximately \(\boxed{0.0780}\).