ACMS 20340 Statistics for Life Sciences Chapter 11: The Normal - - PowerPoint PPT Presentation

ACMS 20340 Statistics for Life Sciences

Chapter 11: The Normal Distributions

Introducing the Normal Distributions

The class of Normal distributions is the most widely used variety of continuous probability distributions. Normal density curves are symmetric, single-peaked, and bell-shaped. The are not “normal” in the sense of “typical” or “boring”, but they are actually quite special.

Why “Normal”?

In 1809 Carl Friedrich Gauss developed his “normal law of errors” to help rationalize the use of the method of least squares.

Why “Normal”?

Many years ago I called the Laplace-Gaussian curve the normal curve, which ... has the disadvantage of leading people to believe that all other distributions of frequency are in one sense or another ‘abnormal’.

• Karl Pearson (1920)

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Francis Galton’s Bean Machine

The first generator of Normal random variables.

The Shape of Normality

The Shape of Normality

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The density curve of a particular Normal distribution is described by its mean µ and its stand deviation σ. The equation of the density curve is f (x) = 1 √ 2π e− 1

2( x−µ σ ) 2

.

Mean and Standard Deviation

Changing the mean, µ, merely changes where the curve is centered.

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5 6 7 8 9 :5 :6 :7 :8 :9 65 66 67 68 69 ;5

Here are Normal curves with µ = 10, 15, and 20, and σ = 3.

Mean and Standard Deviation

Changing the standard deviation, σ, changes the spread of the curve. This also changes the height (since the area = 1).

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Here are curves with µ = 15 and σ = 2, 4, and 6.

Why Care About Normal Distributions?

• 1. They provide good descriptions of real data, including many

biological characteristics, such as blood pressure, bone density, heights, and yields of corn.

• 2. They provide good approximations of many chance outcomes,

such as the proportion of boys over many hospital births.

• 3. Many statistical inference methods rely on Normal
• distributions. (We’ll see this in chapter 13 and beyond.)

Warning!

Do not assume that every variable has a Normal distribution!! For example, the guinea pig survival times are skewed to the right.

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A Few Words on Notation

There is a common shorthand for Normal distributions. A Normal distribution with mean µ and standard deviation σ is denoted N(µ, σ).

The 68-95-99.7 Rule

For a Normal distribution with mean µ and standard deviation σ, (i.e. N(µ, σ)),

◮ approximately 68% of observations fall within σ of µ; ◮ approximately 95% of observations fall within 2σ of µ; and ◮ approximately 99.7% of observations fall within 3σ of µ.

This rule holds for all Normal distributions.

The 68-95-99.7 Rule

Heights of Young Women

The heights of young women between ages 18 to 24 are approximately Normally distributed with µ = 64.5 in. and σ = 2.5 in.

!

Heights of Young Women

What is the probability that a young woman is taller than 62 inches? (% between 62 and 67) + (% above 67) = (% above 62) 68% + 16% = 84%

Standard Normal Distribution

There are many possible Normal distributions (one for every µ and positive σ). By sliding and stretching the curve, we can transform any Normal distribution to any other Normal distribution. Not only do Normal distributions share common properties. We single out the Normal distribution N(0, 1), and call it the standard Normal distribution. And we call the transformation of an arbitrary Normal distribution to the standard one standardizing.

Standardizing

If x is an observation from the Normal distribution N(µ, σ), the standardized value of x is z = x − µ σ The standardized values are often called z-scores.

z-scores

z-scores measure how many standard deviations an observation is away from the mean. A positive z-score indicates the observation is greater than the mean. A negative z-score indicates the observation is less than the mean.

Heights of Young Women

Recall the height distribution of young women is N(64.5, 2.5). The standardized height is z = height − 64.5 2.5

◮ A woman 70 inches tall has the z-score

z = 70 − 64.5 2.5 = 2.2.

◮ A woman 5 feet (60 inches) tall has the z-score

z = 60 − 64.5 2.5 = −1.8.

Finding Normal Probabilities

Whether using software or tables, Normal probabilities are given as cumulative probabilities. The cumulative probability for a value x is the proportion of

• bservations less than or equal to x.

Tips for Finding Normal Probabilities

We use the addition rule and the complement rule to find probabilities. Recall that the probability of any individual value is 0. So, P(X ≤ 40) = P(X < 40) + P(X = 40) = P(X < 40). Sketching a picture of the area you want can be very helpful.

Methods of Finding Normal Probabilities

◮ Normal Curve applet on the website ◮ CrunchIt! distribution calculator ◮ Standard Normal Tables

The Standard Normal Table

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The Standard Normal Table

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.0062 is the area under N(0,1) left of z = -2.50

The Standard Normal Table

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.0062 is the area under N(0,1) left of z = -2.50 .0060 is the area under N(0,1) left of z = -2.51

The Standard Normal Table

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.0062 is the area under N(0,1) left of z = -2.50 .0060 is the area under N(0,1) left of z = -2.51 .0052 is the area under N(0,1) left of z = -2.56

Heights of Young Women 1

What is the probability that a randomly selected young woman measures between 60 and 68 inches tall? Recall, our distribution is N(64.5, 2.5). First, let’s sketch the density curve to find the cumulative probabilities we need.

Heights of Young Women

We need to find the area less than 68 inches and subtract the area less than 60 inches.

Heights of Young Women 2

First we find the z-scores by standardizing: If 60 ≤ x < 68, then it follows that 60 − 64.5 2.5 ≤ x − 64.5 2.5 < 68 − 64.5 2.5 . Thus if we set z = x−64.5

2.5

, we have −1.8 ≤ z < 1.4.

Heights of Young Women 3

Second, we use the table to find the areas.

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Heights of Young Women 4

Lastly, we finish the calculation. Area between -1.8 and 1.4 = (area left of 1.4) − (area left of -1.8) = 0.9192 − 0.0359 = 0.8833 The probability that a randomly selected young woman measures between 60 and 68 inches tall is about 0.88 or 88%.

Question 7

Using N(1, 5), for what value of a do we have P(X < a) = 1

4?

Standardize a: za = a − 1 5 Use table to find za such that P(Z < za) = 1

4.

P(Z < −0.68) = 0.2483 P(Z < −0.67) = 0.2514 Choose za = −0.67 since that is closer to 0.25. Now solve for a: za = −0.67 = a − 1 5 = ⇒ a = (5)(−0.67) + 1 = −2.35

The Other Quartile

Using N(1, 5), for what value of a do we have P(X < a) = 1

4?

a = −2.35 Now, for what value of a do we have P(X > a) = 0.25? P(X < a) = 1 − 0.25 = 0.75. za = a − 1 5 The table gives P(Z < 0.67) = 0.75. a = (5)(0.67) + 1 = 4.35

General Procedure

To solve P(X < a) = p for some normal distribution N(µ, σ):

◮ Find the corresponding za value using the table for the

standard normal distribution: P(Z < za) = p.

◮ Use algebra to solve for a: a = µ + σza.