Answer :
Sure! Let's solve this problem step by step. We have a population with a normal distribution where the mean ([tex]\(\mu\)[/tex]) is 174.6 and the standard deviation ([tex]\(\sigma\)[/tex]) is 99.9. We'll address both parts of the question.
### Part a: Probability that a single randomly selected value is greater than 148.2
1. Find the z-score:
The z-score is a measure of how many standard deviations an element is from the mean. It is calculated using the formula:
[tex]\[
z = \frac{X - \mu}{\sigma}
\][/tex]
where [tex]\(X\)[/tex] is the value of interest (148.2), [tex]\(\mu\)[/tex] is the mean (174.6), and [tex]\(\sigma\)[/tex] is the standard deviation (99.9).
2. Calculate the z-score:
[tex]\[
z = \frac{148.2 - 174.6}{99.9}
\][/tex]
3. Find the probability:
Once we have the z-score, we use the standard normal distribution (z-table) to find the probability that a value is less than 148.2, and then subtract it from 1 to find the probability that a value is greater than 148.2.
[tex]\[
P(X > 148.2) = 1 - P(Z \leq z)
\][/tex]
4. Result:
The probability that a single randomly selected value is greater than 148.2 is approximately 0.6042.
### Part b: Probability that a randomly selected sample of size [tex]\(n=112\)[/tex] has a mean greater than 148.2
1. Calculate the Standard Error (SE):
The standard error of the mean is calculated using:
[tex]\[
SE = \frac{\sigma}{\sqrt{n}}
\][/tex]
where [tex]\(\sigma\)[/tex] is the standard deviation (99.9), and [tex]\(n\)[/tex] is the sample size (112).
2. Calculate the z-score for the sample mean:
[tex]\[
z = \frac{X - \mu}{SE}
\][/tex]
3. Find the probability:
Similarly to part a, find the probability that the sample mean is greater than 148.2 using the z-score.
[tex]\[
P(M > 148.2) = 1 - P(Z \leq z)
\][/tex]
4. Result:
The probability that the mean of a randomly selected sample of size 112 is greater than 148.2 is approximately 0.9974.
By following these steps, we determine the probabilities for both a single value and a sample mean from this normally distributed population.
### Part a: Probability that a single randomly selected value is greater than 148.2
1. Find the z-score:
The z-score is a measure of how many standard deviations an element is from the mean. It is calculated using the formula:
[tex]\[
z = \frac{X - \mu}{\sigma}
\][/tex]
where [tex]\(X\)[/tex] is the value of interest (148.2), [tex]\(\mu\)[/tex] is the mean (174.6), and [tex]\(\sigma\)[/tex] is the standard deviation (99.9).
2. Calculate the z-score:
[tex]\[
z = \frac{148.2 - 174.6}{99.9}
\][/tex]
3. Find the probability:
Once we have the z-score, we use the standard normal distribution (z-table) to find the probability that a value is less than 148.2, and then subtract it from 1 to find the probability that a value is greater than 148.2.
[tex]\[
P(X > 148.2) = 1 - P(Z \leq z)
\][/tex]
4. Result:
The probability that a single randomly selected value is greater than 148.2 is approximately 0.6042.
### Part b: Probability that a randomly selected sample of size [tex]\(n=112\)[/tex] has a mean greater than 148.2
1. Calculate the Standard Error (SE):
The standard error of the mean is calculated using:
[tex]\[
SE = \frac{\sigma}{\sqrt{n}}
\][/tex]
where [tex]\(\sigma\)[/tex] is the standard deviation (99.9), and [tex]\(n\)[/tex] is the sample size (112).
2. Calculate the z-score for the sample mean:
[tex]\[
z = \frac{X - \mu}{SE}
\][/tex]
3. Find the probability:
Similarly to part a, find the probability that the sample mean is greater than 148.2 using the z-score.
[tex]\[
P(M > 148.2) = 1 - P(Z \leq z)
\][/tex]
4. Result:
The probability that the mean of a randomly selected sample of size 112 is greater than 148.2 is approximately 0.9974.
By following these steps, we determine the probabilities for both a single value and a sample mean from this normally distributed population.
