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.

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).

If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track.

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Normal Probability Practice Problems and Answers was last modified: October 12th, 2017 by Stephanie

6 thoughts on “Normal Probability Practice Problems and Answers”

Lance R.

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.

Andale Post author

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

James T. Young

1. 99.38%
2. 30.85%
3. 90.82%
4. 15.87%
5. 84/13%

N C Zulu

I need picture of it and answers

Derek Brooks

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%.

Andale Post author

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?

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?