Assume that the average score of the Final exam of a particular class is μ = 82.0 and standard deviation σ = 15. The probability (proportion) of a student selected at random to have a score more than 90 on the exam is approximately:

Question

A) 0.296
B) 0.704
C) 0.180
D) 0.819
E) None of the above

Answer 1

To find the probability that a randomly selected student scores more than 90 on the exam, we can use the properties of the normal distribution.

Given:

• Mean (μ) = 82.0
• Standard deviation (σ) = 15.0
• Score (X) = 90

We can standardize the score to find the corresponding z-score using the formula:

$$z=\frac{X-\mu }{\sigma }$$

Substituting the values:

$$z=\frac{90-82}{15}=\frac{8}{15}\approx 0.5333$$

Next, we will look up the z-score of approximately 0.53 in the standard normal distribution table, or we can use a calculator. The z-table tells us the probability that a value is less than a certain z-score.

For $z=0.53$, the cumulative probability (P(Z < 0.53)) is approximately 0.7027.

To find the probability of a score greater than 90, we calculate:

$$P(X>90)=1-P(Z<0.53)=1-0.7027\approx 0.2973$$

Rounding this result, we find:

$$P(X>90)\approx 0.296$$

Therefore, the correct answer is:

A) 0.296