Normal DistributionZ-scores put to use!

Section 2.2

Reference Text:

The Practice of Statistics, Fourth Edition.

Starnes, Yates, Moore

Today’s Objectives

• The 68-95-99.7 Rule

• State mean an standard deviation for The Standard Normal Distribution

• Given a raw score from a normal distribution, find the standardized “z-score”

• Use the Table of Standard Normal Probabilities to find the area under a given section of the Standard Normal curve.

The 68-95-99.7 Rule

• How many standard deviations do you think it would take for us to have the entire sample or population accounted for and just have a .03% uncertainty?

• In other words, how many standard deviations away from the mean encompasses almost all objects in the study?

The 68-95-99.7 Rule

• 3!• The 68-95-99.7 Rule describes the percent of

observations fall within 1,2 or 3 standard deviations. Look at the visual:

The 68-95-99.7 Rule

• So, – Approximately 68% of the observations fall

within of the mean µ – Approximately 95% of the observations fall

within 2 of the mean µ – Approximately 99.7% of the observations fall

within 3 of the mean µ

The 68-95-99.7 Rule

• If I have data within 2 standard deviations, then I'm accounting for 95% of observations

• Question: what percent is in the left tail?

You Try!

• The distribution of number of movies AP Statistic students watch in two weeks is close to normal. Suppose the distribution is exactly Normal with mean µ= 6.84 and standard deviation = 1.55 (this is non fiction data)

• A) Sketch a normal density curve for this distribution of movies watched. Label the points that are one, two, and three SD away from the mean.

• B) What percent of the movies is less that 3.74? Show your work!• C) What percent of scores are between 5.29 and 9.94? Show work!• Remember: Always put your answers back into context!

Break!

- 5 Minutes

Standardizing Observations• All normal distributions have fundamentally the same

shape.• If we measure the x axis in units of size σ about a

center of 0, then they are all exactly the same curve.• This is called the Standard Normal Curve

– We abbreviate the normal dist. As N( µ, )

• To standardize observations, we change from x values (the raw observations) z values (the standardized observations) by the formula:

xz

The Standard Normal Distribution

• Notice that the z-score formula always subtracts μ from each observation.– So the mean is always shifted to zero

• Also notice that the shifted values are divided by σ, the standard deviation.– So the units along the z-axis represent

numbers of standard deviations

• Thus the Standard Normal Distribution is always N(0,1).

Example!• The heights of young women are:

N(64.5, 2.5)

• Use the formula to find the z-score of a woman 68 inches tall.

• A woman’s standardized height is the number of standard deviations by which her height differs from the mean height of all young women.

68 64.51.4

2.5z

Normal Distribution Calculations• What proportion of all young women are less

than 68 inches tall? – Notice that this does not fall conveniently on one of the σ

borders

– We already found that 68 inches corresponds to a z-score of 1.4

• So what proportion of all standardized observations fall to the left of z = 1.4?

• Since the area under the Standard Normal Curve is always 1, we can ask instead, what is the area under the curve and to the left of z=1.4– For that, we need a table!!

The Standard Normal Table• Find Table A of the handout

– It is also in your textbook in the very back

• Z-scores (to the nearest tenth) are in the left column– The other 10 columns round z to the nearest hundredth

• Find z = 1.4 in the table and read the area– You should find area to the left = .9192

• So the proportion of observations less than z = 1.4 is about 92%– Now put the answer in context: “About 92% of all

young women are 68 inches tall or less.”

What about area above a value?

• Still using the N(64.5, 2.5) distribution, what proportion of young women have a height of 61.5 inches or taller?

• Z = (61.5 – 64.5)/2.5 = -1.2

• From Table A, area to the left of -1.2 =.1151– So area to the right = 1 - .1151 = .8849

• So about 88.5% of young women are 61.5” tall or taller.

What about area between two values?

• What proportion of young women are between 61.5” and 68” tall?

• We already know 68” gives z = 1.4 and area to the left of .9192

• We also know 61.5” gives z = -1.2 and area to the left of .1151

• So just subtract: .9192 - .1151 = .8041• So about 80% of young women are between

61.5” and 68” tall– Remember to write your answer IN CONTEXT!!!

Given a proportion, find the observation x

• SAT Verbal scores are N(505, 110). How high must you score to be in the top 10%?

• If you are in the top 10%, there must be 90% below you (to the left).

• Find .90 (or close to it) in the body of Table A. What is the z-score?– You should have found z = 1.28

• Now solve the z definition equation for x

• So you need a score of at least 646 to be in the top 10%.

5051.28

1101.28 110 505

645.8

xz

x

x

x

How to Solve Problems Involving Normal Distribution

• State: Express the problem in terms of the observed variable x

• Plan: draw a picture of the distribution and shade the area of interest under the curve.

• Do: Preform the calculations– Standardize x to restate the problem in terms of standard normal

variable z– Use Table A and the fact that the total area under the curve is 1

to find the required area under the standard normal curve

• Conclude: Write your conclusion in context of the problem.

• Lets look at TB pg 120 “Tiger on the Range”

Today’s Objectives

• The 68-95-99.7 Rule

• State mean an standard deviation for The Standard Normal Distribution

• Given a raw score from a normal distribution, find the standardized “z-score”

• Use the Table of Standard Normal Probabilities to find the area under a given section of the Standard Normal curve.

Homework

TB Pg 131: 41-74 (multiples of 3)