Z-Scores and Z-Score Table Quiz

The mean number of accidents a week at a company is 6.4 with a standard deviation of 1.5. What proportion of weeks would you expect to have less than 5 accidents?

Hint: Find z-score, then look up z-score in table

• 0.6915

• 0.1762

• 0.8238

• -0.93

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The 40 yards sprint times for a soccer team are found to be normally distributed with a mean of 5.2 seconds and a standard deviation of 0.3 seconds. What is the z-score for a player who runs a time of 5.6 seconds?

• -1.33

• 1.33

• 0.88

• 1.02

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CCSS.HSS.ID.A.4

The average waist size for teenage males is 29 inches with a standard deviation of 1.4 inch.

If waist sizes are normally distributed, estimate the proportion of teenagers who will have waist sizes greater than 31 inches?

Hint: Find z-score then look up z-score in table and convert to a percent

• 92.4%

• 10.5%

• 16%

• 7.6%

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The scoring of modern IQ is such that Intelligence Quotients (IQs) have a normal distribution of μ= 100 and 𝞂 = 15.

Mensa International is a non-profit organization that accepts only people with IQ score within the top 2%. What level of IQ qualifies one to be a member of Mensa?

Hint: Work backwards...convert percent to a decimal and find decimal in z-score table chart to get z-score and then use z-score to find IQ value

• 115

• 130.75

• 145

• 120

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The distances male long jumpers for State College jump are approximately normal with a mean of 263 inches and a standard deviation of 14 inches.

Suppose a male long jumper's jump ranked in the 75th percentile. How long was his jump?

Hint: work backwards... change percentile to decimal and find within z-score chart to find z-score and then convert z-score to long jump distance

• 0.67 inch

• 263 inches

• 272.44 inches

• 277.25 inches

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CCSS.HSS.ID.A.4

The shelf life of a particular dairy product is normally distributed with a mean of 12 days and a standard deviation of 3 days.

About what percent of the products last between 12 and 15 days?

Hint: Find the z-score of each value, subtract and look up value in z-score table and convert to percent or use empirical rule

• 68%

• 34%

• 16%

• 2.5%

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Find the z-score for 48 given a mean of 50 and a standard deviation of 5.

• 2.5

• -2.5

• -0.4

• 0.4

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Find the percent under the standard normal curve that lies to the left of z=2.1.

Hint: Use z-score table and convert to percent

• 1.79%

• -1.79%

• 98.21%

• 0.98%

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Find the percent under the standard normal curve that lies to the right of z=-0.67.

Hint: find the value in the z-score table and convert to %

• 63.42%

• 74.86%

• 25.14%

• 50%

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CCSS.HSS.ID.A.4

A test has a mean of 40 and a standard deviation of 10. What is the percent of people that scored above 62?

Hint: Find the z-score and look up the value in the z-score table and convert to a percent

• 1.39%

• 45.34%

• 98.41%

• 22.34%

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