To find the required probabilities, we will use the properties of the normal distribution and the z-score formula.
- Finding the probability that a single randomly selected value is less than 97 (P(X < 97)):
Given:
- Mean (μ) = 65.4
- Standard deviation (σ) = 61.9
- Value (X) = 97
We first calculate the z-score, which is given by the formula:
\[ z = \frac{X - \mu}{\sigma} \]
Substituting the values:
\[ z = \frac{97 - 65.4}{61.9} = \frac{31.6}{61.9} \approx 0.5097 \]
Now, we will look up the z-score in the standard normal distribution table or use a calculator to find the probability associated with this z-score.
Using a z-table or calculator:
\[ P(X < 97) \approx P(Z < 0.5097) \approx 0.6943 \]
- Finding the probability that a sample of size n=24 has a mean less than 97 (P(M < 97)):
When dealing with sample means, we need to consider the standard error (SE), which is calculated as:
\[ SE = \frac{\sigma}{\sqrt{n}} \]
Where:
- n = 24
- σ = 61.9
Calculating the standard error:
\[ SE = \frac{61.9}{\sqrt{24}} \approx \frac{61.9}{4.899} \approx 12.629 \]
Next, we calculate the z-score for the sample mean using the same formula as before:
\[ z = \frac{M - \mu}{SE} \]
Substituting the values:
\[ z = \frac{97 - 65.4}{12.629} = \frac{31.6}{12.629} \approx 2.5043 \]
Now, we find the probability associated with this z-score:
\[ P(M < 97) \approx P(Z < 2.5043) \]
Using a z-table or calculator:
\[ P(Z < 2.5043) \approx 0.9939 \]
- Final Answers:
Therefore,
- \( P(X < 97) \approx 0.6943 \)
- \( P(M < 97) \approx 0.9939 \)
Final answers (accurate to 4 decimal places):
- P(X < 97) = 0.6943
- P(M < 97) = 0.9939