Finding Proportions
Calculating Z-Score
Finding Observations
Percentile and Interpretations
Finding Mean or SD
100

Find the proportion of observations from a Standard Normal distribution if z<-1.18


Round at least 2 places past the decimal (hundredths or thousandths)

0.119

100

Suppose the following heights in inches is N(64.6, 2.15). What is the z-score of someone who is 68 inches? Round to the nearest hundredth.

1.58

100

Suppose the distribution of heights in inches is N(66.4, 2.89). At what height is the 10th percentile? Round to the nearest hundredth.

62.70

100
A person is in the 10th percentile for their height. What does this mean?

She is taller than 10% of people.


Or 10% of people are shorter than her.

100

Suppose the 20th percentile of heights is 63.4 inches and the 80th percentile is 70.2. What is the mean of this distribution?

66.8 inches

200

Find the proportion of observations from a Standard Normal distribution if -2.05<z<2.08


Round at least 2 places past the decimal (hundredths or thousandths)

0.961

200

Suppose the following heights in inches is N(64.6, 2.15). What is the z-score of someone who is 52 inches? Round to the nearest hundredth.

-5.86

200

Suppose the distribution of heights in inches is N(66.4, 2.89). At what height is the 90th percentile? Round to the nearest hundredth.

70.10 inches

200

What is a z-score?

It tells you how many standard deviations above or below the mean you are.

200

Suppose the 20th percentile of heights is 63.4 inches and the 80th percentile is 70.2. What is the standard deviation of this distribution?

SD = 4.04

Rounding might be off

300

Suppose the following heights in inches is N(64.6, 2.15). Find the probability that someone is taller than 68 inches.


Round to the nearest hundredth or thousandth.

0.057

300

Someone is in the 20th percentile in their class in terms of height. What is their z-score? Round to the nearest hundredth.

-0.84

300

Suppose the distribution of heights in inches is N(66.4, 2.89). How tall do you have to be to be in the top 20th percentile? Round to the nearest hundredth.

68.83

300

A person's height is 63 inches and in this distribution, this yields a z-score of -2.1. Interpret the z-score in this context.

She is 2.1 standard deviations below the mean.


Or: Her height is 2.1 standard deviations below the mean.

300

Suppose the 73rd percentile of heights is 68.3 inches and the 27th percentile is 62.1. What is the standard deviation of this distribution?

Round to the nearest hundredth. Show all work.

SD: 5.06 (rounding might be slightly off)

400

Suppose the following heights in inches is N(64.6, 2.15). Find the probability that someone is between 62.5 and 72.1 inches.


Round to the nearest hundredth or thousandth.

0.835

400

Someone is in the 99th percentile in their class in terms of height. What is their z-score? Round to the nearest hundredth.

2.33

400

Suppose the distribution of heights in inches is N(66.4, 2.89). At what heights are the middle 80th percentile?

62.70 and 70.10 inches

400

Suppose you surveyed 15 people. 3 are 62 inches, 4 are 69 inches, and 8 are 66 inches. Interpret the percentile of someone who is 66 inches.

Someone who is 66 inches is in the 26.67th percentile, which means they are taller than 26.67% of the students in this distribution (or 26.67% of students are shorter than them).
400

Suppose the 3rd percentile of heights is 59.55 inches and the 87th percentile is 65.87 What is the mean of this distribution?

Round to the nearest hundredth. Show all work.

Mean = 63.5

Rounding might be a bit off.

500

Suppose the following heights in inches is N(64.6, 2.15). Find the probability that someone is shorter than 50 inches.


Round to the nearest hundredth or thousandth.

0
500

Someone is in the 22th percentile in their class in terms of height. Another person is in the 81st percentile. What are their z-scores? Round to the nearest hundredth.

-0.77 and 0.88

500

Suppose the distribution of heights in inches is N(66.4, 2.89). At what heights are the middle 60th percentile?

63.97 and 68.83 inches

500

Suppose the distribution of heights in inches is N(66.4, 2.89). Interpret the percentile of someone who is 67 inches.

Probability is 0.5822 which means they are taller than 58.22% of people in this distribution.

500

Suppose the 3rd percentile of heights is 59.55 inches and the 87th percentile is 65.87 What is the standard deviation of this distribution?

Round to the nearest hundredth. Show all work.

SD = 2.1

Rounding might a bit off.

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