Probability and Statistics > Probability > Normal Probability Practice Problems

These normal probability practice problems will help you practice calculating z-scores and using the z-table.

General steps for solving normal probability practice problems:

- Use the equation above to find a z-score. If you don’t know how to look up z-scores (or if you want more practice, see this z-score article for videos and step-by-step instructions.
- Look up the z-score in the z-table and find the area.
- Convert the area to a percentage.

If you more need help with these steps, see the normal distribution word problems index, which outlines the steps in detail (with videos).

## Normal Probability Practice Problems.

**Click on the question for the answer!**

**1. Scores on a particular test are normally distributed with a standard deviation of 4 and a mean of 30. What is the probability of anyone scoring less than 40?**

- z = 40-30/4 = 2.5. Area to the left of 2.5 =.9938.

**2. Annual salaries for a large company are approximately normally distributed with a mean of $50,000 and a of $20,000. What percentage of company workers may under $40,000?**

- z = 40000-50000/20000 = -.5 = .3085 = 30.85%.

**3. IQ scores have a normal distribution with a mean of 100 and a standard deviation of 15. What percent of people have an IQ above 120?**

- z = (120-100)/15 = 1.33 = 1 – 0.9082 = 0.0918 (9.18%).

**4. The amount of time a student taking statistics spends on studying for a test is normally distributed. If the average time spent studying is 12 hours and the standard deviation is 4 hours, what is the probability that a student will spend more than 8 hours studying?**

- z = (8-12) / 4 = 1 – .1587 = 0.8413 = 84.13%

**5. The amount of candy dispensed by a candy machine is normally distributed with a mean of 0.9 oz and a standard deviation of 0.1 ounces. If the machine is used 500 times, how many times will it dispense more than 1 oz of candy?**

- First, find the probability of more than 1 oz candy being dispensed: z = 1 – 0.9 / 0.1 = 1 = .8413. The area greater than this (to the right of the z-score) is 1 – .8413 = 0.1587. Finally, find 15.87% of 500 = 79.35 (rounded down that’s 79 times out of 500).

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About question 5, the probability associated with the Z score of 1 is 0.84, and we’re looking for the chance of the machine dispensing greater than 1oz candy. Shouldn’t we take 1-0.84 and multiply that probability to obtain an expected number of candy>1oz?

I’m confused about the solution provided for this problem. Hope you can clarify.

Lance,

You are completely right. A step was missing from the end. That is now fixed.

Regards,

S

1. 99.38%

2. 30.85%

3. 90.82%

4. 15.87%

5. 84/13%

I need picture of it and answers

Number 3’s answer is incorrect. If we want above 120, you need to subtract .9082 from 1. The answer should be .0918 or 9.18%. Same logic for problem 4. We want more than 8 hours of studying. Subtract .1587 from 1. The answer is 84%.

Hi, Derek,

Thanks for commenting. I’m actually seeing those as the correct answers (for example, the solution shows for q3 is .9.18%. ) — not sure where you are seeing different numbers?