To calculate the probabilities you've requested, we need to use the properties of the normal distribution.
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Finding P(X > 33.2):
We are looking for the probability that a single randomly selected value (X) is greater than 33.2. Given that the population mean (μ) is 21 and the standard deviation (σ) is 33.7, we first compute the z-score for X = 33.2.
The formula for the z-score is: \[ z = \frac{X - \mu}{\sigma} \]
Plugging in the values: \[ z = \frac{33.2 - 21}{33.7} = \frac{12.2}{33.7} \approx 0.362 \]
Next, we look up the z-score in the standard normal distribution table or use a calculator to find P(Z > 0.362).
The cumulative probability for Z = 0.362 is approximately 0.6438, so: \[ P(X > 33.2) = 1 - P(Z \leq 0.362) = 1 - 0.6438 = 0.3562 \]
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Finding P(M > 33.2) for a sample of size n = 15:
When sampling from a population, the distribution of the sample mean (M) will also be normally distributed with mean μ and standard deviation (standard error) calculated as follows: \[ \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{33.7}{\sqrt{15}} \approx \frac{33.7}{3.872} \approx 8.692 \]
Now we again calculate the z-score for M = 33.2: \[ z = \frac{M - \mu}{\sigma_M} = \frac{33.2 - 21}{8.692} = \frac{12.2}{8.692} \approx 1.402 \]
We now find the probability: \[ P(M > 33.2) = 1 - P(Z \leq 1.402) \]
Checking the z-table for P(Z ≤ 1.402), we find it approximately equals 0.9192. Therefore: \[ P(M > 33.2) = 1 - 0.9192 = 0.0808 \]
So, rounding to four decimal places, our results are:
- \( P(X > 33.2) \approx 0.3562 \)
- \( P(M > 33.2) \approx 0.0808 \)
Final answers:
- \( P(X > 33.2) = 0.3562 \)
- \( P(M > 33.2) = 0.0808 \)