Question

Entry to a certain university is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100. Tom wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Tom takes the test and scores 585. What percentage did Tom get?

Answer 1

To find out what percentage of students Tom scored better than, you can use the z-score formula:

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

Where:
- \(X\) is Tom's score (585)
- \(\mu\) is the mean score (500)
- \(\sigma\) is the standard deviation (100)

First, calculate the z-score:

\[Z = \frac{585 - 500}{100} = 0.85\]

Now, you want to find the percentage of students who scored worse than Tom, which means you need to find the cumulative probability up to the z-score of 0.85. You can look up this value in a standard normal distribution table or use a calculator. The cumulative probability for a z-score of 0.85 is approximately 0.8023.

To find the percentage, subtract this cumulative probability from 1 and then multiply by 100:

\[(1 - 0.8023) * 100 \approx 19.77\%\]

So, Tom scored better than approximately 19.77% of the students who took the test.