To solve both parts of your question, we will use the properties of the normal distribution.
Part 1: Probability of a single randomly selected value being greater than 33.2
Given:
- Mean (\( \mu \)) = 21
- Standard deviation (\( \sigma \)) = 33.7
We first need to find the z-score for the value 33.2 using the formula: \[ z = \frac{X - \mu}{\sigma} \] Where \( X \) is the value we are interested in.
Substituting the values: \[ z = \frac{33.2 - 21}{33.7} = \frac{12.2}{33.7} \approx 0.3626 \]
Now, we will use the z-score to find the probability. We look up the z-score in the standard normal (Z) table or use a calculator to find \( P(Z > 0.3626) \).
From Z-tables or calculators: \[ P(Z < 0.3626) \approx 0.6430 \] Thus, \[ P(Z > 0.3626) = 1 - P(Z < 0.3626) \approx 1 - 0.6430 = 0.3570 \]
So, the probability that a single randomly selected value is greater than 33.2 is: \[ P(X > 33.2) \approx 0.3570 \]
Part 2: Probability that the sample mean of size n=15 is greater than 33.2
For the sample mean, we will use the sampling distribution of the mean. The mean of the sample mean distribution is the same as the population mean (\( \mu_M = \mu = 21 \)), and the standard deviation of the sample mean (standard error) is given by: \[ \sigma_M = \frac{\sigma}{\sqrt{n}} \] Where \( n = 15 \).
Calculating the standard error: \[ \sigma_M = \frac{33.7}{\sqrt{15}} \approx \frac{33.7}{3.873} \approx 8.695 \]
Now we will find the z-score for the sample mean of 33.2: \[ z = \frac{M - \mu_M}{\sigma_M} = \frac{33.2 - 21}{8.695} = \frac{12.2}{8.695} \approx 1.403 \]
Now we find \( P(M > 33.2) = P(Z > 1.403) \).
Using the Z-tables or a calculator: \[ P(Z < 1.403) \approx 0.9207 \] Thus, \[ P(Z > 1.403) = 1 - P(Z < 1.403) \approx 1 - 0.9207 = 0.0793 \]
So, the probability that a sample of size n=15 is randomly selected with a mean greater than 33.2 is: \[ P(M > 33.2) \approx 0.0793 \]
Summary of Results
- \( P(X > 33.2) \approx 0.3570 \)
- \( P(M > 33.2) \approx 0.0793 \)
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