To find the probability that a randomly selected student has a score greater than 90 on the exam, we will use the standard normal distribution (Z-score).
- First, we calculate the Z-score for a score of 90 using the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
- \( X \) is the score we are interested in (90),
- \( \mu \) is the mean (82),
- \( \sigma \) is the standard deviation (15).
Substituting the values:
\[ Z = \frac{90 - 82}{15} = \frac{8}{15} \approx 0.5333 \]
- Next, we look up the Z-score of approximately 0.5333 in the standard normal distribution table (or use a calculator with normal distribution functions). The value we find will give us the probability of a student scoring less than 90.
The Z-table or calculator usually shows that:
\[ P(Z < 0.5333) \approx 0.703 \]
- To find the probability of a student scoring more than 90, we subtract this probability from 1:
\[ P(Z > 0.5333) = 1 - P(Z < 0.5333) \approx 1 - 0.703 \approx 0.297 \]
Thus, the probability (proportion) of a student selected at random to have a score more than 90 on the exam is approximately:
A) 0.296.
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