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**#**11.15 – 11.24, 11.28, 11.30, 11.31, 11.35

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The Normal Distribution

Chapter 11

Normal Curves

• Normal curves are symmetric, single-peaked, and bell-shaped distributions.

• The exact density curve for a particular Normal distribution is described by giving its

mean, 𝜇, and its standard deviation, 𝜎.

• The mean of a Normal distribution is at the center, and the standard deviation is the

distance from the center to the change-of-curvature points on either side.

Not all data are

Normally distributed.

𝑓 𝑥 =

1

2𝜋

1 𝑥−𝜇 2

−2 𝜎

𝑒

Here means are the same (µ = 15)

whereas standard deviations are

different (σ = 2, 4, and 6).

Here means are different

(µ = 10, 15, and 20) whereas standard

deviations are the same (σ = 3).

Human heights, by gender, can be modeled

quite accurately by a Normal distribution.

Guinea pigs survival times after inoculation of a

pathogen are clearly not a good candidate for a

Normal model!

The 68-95-99.7 Rule (Empirical Rule)

All normal curves, 𝑁(µ, 𝜎), share the same properties:

•

•

•

•

About 68% of all observations are within 1 standard deviation (σ) of the mean (µ).

About 95% of all observations are within 2 σ of the mean µ.

Almost all (99.7%) observations are within 3 σ of the mean.

To obtain any other area under a Normal curve, use either technology or Table B.

Example

The distribution of heights of young women age 18 to 24 is approximately normal

with 𝜇 = 64.5inches and standard deviation of 𝜎 = 2.5inches. According to the

Empirical Rule, the heights of the middle 95% of young women are between what

two values?

𝜇 − 2 𝜎 = 64.5 − 2 2.5 = 64.5 − 5 = 59.5 𝑖𝑛𝑐ℎ𝑒𝑠

𝜇 + 2 𝜎 = 64.5 + 2 2.5 = 64.5 + 5 = 69.5 𝑖𝑛𝑐ℎ𝑒𝑠

The middle 95% of women’s heights is between 59.5 inches and 69.5 inches.

Example

A height of 62 inches is one standard deviation below the mean. What is the

probability that a woman is taller than 62 inches?

The probability that a randomly chosen female between 18 to 24 years

old is taller than 62 inches is approximately 84%.

Z-Score

If 𝑥 is an observation from a distribution that has mean 𝜇 and standard deviation,

𝜎, the standardized value of 𝑥 is:

𝑥−𝜇

𝑧=

𝜎

A standardized value is often called a z-score.

A z-score tells us how

many standard deviations

the original observation

falls away from the mean,

and in which direction.

Example

The heights of young women aged 18 to 24 are approximately Normal, with 𝜇 =

64.5 inches and standard deviation of 𝜎 = 2.5 inches. Determine a woman’s

standardized height if she is 70 inches tall, similarly for a woman who is 60 inches

tall (5ft).

70 − 64.5

𝑧=

= 2.2

2.5

60 − 64.5

𝑧=

= −1.8

2.5

Standard Normal Distribution

The standard Normal distribution is the Normal distribution 𝑁(0,1) with mean 0

and standard deviation 1.

If a variable x has any Normal distribution 𝑁(𝜇, 𝜎) with mean 𝜇 and standard

deviation 𝜎, then the standardized variable

𝑥−𝜇

𝑧=

𝜎

has the standard Normal distribution.

Areas under a Normal curve represent proportions (frequencies) of observations

from that Normal distribution.

The cumulative probability for a value 𝑥 in a distribution is the proportion of

observations in the distribution that lie at or below 𝑥.

Example

Osteoporosis is a condition in which the bones become brittle to the breaking

point due to loss of minerals. To diagnose osteoporosis, bone mineral density

(BMD) is measured and typically reported as a unitless score, which is designed so

that the mean BMD score of healthy young adults of the same sex is zero. The

WHO defining criterion for osteoporosis is a BMD score below -2.5.

Women in their 70s tend to have much lower BMD. In this population, BMD scores

are approximately Normally distributed with a mean of -2 and a standard deviation

of 1. Find the probability that a randomly selected woman in her 70s would have

osteoporosis.

Example

Osteopenia is a condition defined by less than ideal bone density, corresponding to

BMD scores between -2.5 and -1. We want to find what percent of women in their

70s have osteopenia.

53.28% of women in their 70s have

osteopenia.

Percentile

The value such that some percent 𝑝 of the observations in a distribution lie

below it is called the 𝒑th percentile.

Example:

The hatching weights of commercial chickens can be modeled accurately using a

Normal distribution with mean 𝜇 = 45 grams and standard deviation 𝜎 = 4 grams.

What is the third quartile of the distribution of hatching weights?

Example

The lengths of human pregnancies from conception to birth (measured in days)

follow approximately the 𝑁(266,16) distribution. How long are the longest 10% of

pregnancies?

Pregnancies longer than

286 days are in the top

10% of pregnancy

lengths.

Example

The blood cholesterol levels of men aged 55 to 64 are approximately normal

with mean 222 mg/dL and standard deviation 37 mg/dL.

What range of values corresponds to the 10% highest cholesterol levels?

A) > 175 B) > 247

C) > 269

D) > 288

Tricks of Symmetry

Because of the curve’s symmetry, there are two ways of finding the area under

𝑁(0,1) curve to the right of a z-value.

Normal Quintile Plots

• One way to assess if a data set has an approximately Normal distribution is to

plot the data on a Normal quantile plot.

– The data points are ranked and the percentile ranks are converted to zscores. The z-scores are then used for the horizontal axis and the actual

data values are used for the vertical axis.

• If the data have approximately a Normal distribution, the Normal quantile

plot will have roughly a straight-line pattern.

Normal Quintile Plot Examples

Roughly normal

(~ straight-line pattern)

Right skewed

(most of the data points are short

survival times, while a few are longer

survival times)

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