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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].
### 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].
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