Question

Suppose that the test scores of students are normally distributed with the mean 76 and the variance of 64. The probability of students scoring above 82 is?

Answer 1

To find the probability of students scoring above 82, we will first need to standardize the score using the Z-score formula. The Z-score can be calculated as follows:

\[
Z = \frac{X - \mu}{\sigma}
\]

where:
- \(X\) is the value we are interested in (82 in this case),
- \(\mu\) is the mean (76),
- \(\sigma\) is the standard deviation.

Given that the variance is 64, we can find the standard deviation by taking the square root of the variance:

\[
\sigma = \sqrt{64} = 8
\]

Now we can substitute the known values into the Z-score formula:

\[
Z = \frac{82 - 76}{8} = \frac{6}{8} = 0.75
\]

Next, we need to find the probability of a Z-score being greater than 0.75. To do this, we can refer to the standard normal distribution table or use a calculator.

The cumulative probability for \(Z = 0.75\) (which gives the probability of scoring less than 82) is approximately 0.7734. This means that about 77.34% of students scored below 82.

To find the probability of scoring above 82, we subtract this value from 1:

\[
P(X > 82) = 1 - P(X < 82) = 1 - 0.7734 = 0.2266
\]

Thus, the probability of students scoring above 82 is approximately **0.2266**, or **22.66%**.