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------------------------------------------------ A population of values has a normal distribution with [tex]\mu=98.1[/tex] and [tex]\sigma=85.2[/tex]. You intend to draw a random sample of size [tex]n=21[/tex].

1. Find the probability that a single randomly selected value is greater than 38.6.
\[
P(X > 38.6) = \square
\]

2. Find the probability that a sample of size [tex]n=21[/tex] is randomly selected with a mean greater than 38.6.
\[
P(M > 38.6) = 0.9993
\]

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer :

To solve this problem, we need to calculate two probabilities involving a normally distributed population with a mean ([tex]\(\mu\)[/tex]) of 98.1 and a standard deviation ([tex]\(\sigma\)[/tex]) of 85.2. Let's break down each part of the question:

### Part 1: Probability of a Single Value Greater Than 38.6

When dealing with a single value from a normal distribution, we use the z-score formula:

[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]

where [tex]\(X\)[/tex] is the value of interest (38.6 in this case).

1. Calculate the z-score for a single value:

[tex]\[ z = \frac{38.6 - 98.1}{85.2} \][/tex]

2. Find the probability for this z-score:

Convert the z-score to a probability using a standard normal distribution table or a calculator. Since we want the probability that a value is greater than 38.6, we need:

[tex]\[ P(X > 38.6) = 1 - P(X \leq 38.6) \][/tex]

Looking up the cumulative probability for the calculated z-score and subtracting it from 1 gives us the probability we need.

The result is approximately 0.7575.

### Part 2: Probability of the Sample Mean Greater Than 38.6

When evaluating a sample mean, the Central Limit Theorem tells us that the distribution of the sample mean will also be normal with the same mean ([tex]\(\mu = 98.1\)[/tex]) but with a new standard deviation known as the standard error ([tex]\(SE\)[/tex]):

[tex]\[ SE = \frac{\sigma}{\sqrt{n}} \][/tex]

where [tex]\(n = 21\)[/tex] is the sample size.

1. Calculate the standard error:

[tex]\[ SE = \frac{85.2}{\sqrt{21}} \][/tex]

2. Calculate the z-score for the sample mean:

[tex]\[ z = \frac{38.6 - 98.1}{SE} \][/tex]

3. Find the probability for this z-score:

Again, convert the z-score to a probability using a standard normal distribution table or calculator. Since we want the probability that the sample mean is greater than 38.6, we need:

[tex]\[ P(M > 38.6) = 1 - P(M \leq 38.6) \][/tex]

The result is approximately 0.9993.

In conclusion:

- The probability that a single randomly selected value is greater than 38.6 is approximately [tex]\( P(X > 38.6) = 0.7575 \)[/tex].

- The probability that a sample of size 21 has a mean greater than 38.6 is approximately [tex]\( P(M > 38.6) = 0.9993 \)[/tex].