Question

In a math aptitude test, students scores are found to be normally distributed having mean as 45 and standard deviation 5. What percent of students scored more than the mean score?

• A. 45%
• B. 50%
• C. 5%
• D. 60%

Hint:

A normal curve is symmetric about the mean, so half of the area lies above it.

Answer 1

The Correct Option is B

Step 1: Standardise the score.
Convert the mean into a standard normal value: $z = \dfrac{x - \mu}{\sigma} = \dfrac{45 - 45}{5} = 0$.

Step 2: Use the standard normal table.
We need $P(Z > 0)$. The table gives $P(Z \leq 0) = 0.5$, so $P(Z > 0) = 1 - 0.5 = 0.5$.

Step 3: Convert to percent.
$0.5 \times 100 = 50$ percent of the students scored above the mean.

Step 4: Note on the data.
The values 45 and 5 are not needed for the final number, because the standard normal curve is symmetric about zero. Whatever the mean and spread, half of the scores lie above the mean.

Final Answer:
50% of the students scored more than the mean. \[\boxed{50\%}\]