Chapter 6

The Normal Distribution

The Normal Distribution

A continuous, symmetric, bell-shaped distribution of a variable.

Normal Distribution Curve

Finding the area under the Curve

To the left of z– Chart #

To the right of z– 1 – chart #

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Finding the area under the Curve

Between two z scores– Bigger z chart # - smaller z chart #

Tails of two z scores– 1- (Bigger z chart # - smaller z chart #)

Between Zero and #– -Z: .5 – Chart #

- +Z: Chart # - .5

Find the z score when given a percent

The rounding rule

–Z scores are rounded two decimal places

Find the z score when given a percent

To the left:– Find percent in chart then find z score

To the right– 1- given percent, then use chart

Between two z’s– .5- percent/2, then chart

Tails:– Percent/2, then chart

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Using TI-83 Plus

To the left:– invNorm(percent)

To the right– invNorm(1- given percent)

Between two z’s– invNorm(.5- percent/2)

Tails– invNorm(percent/2)

invNorm(

1. Hit 2nd Button

2. Hit DISTR

3. Hit 3 key or arrow down to invNorm

4. Type in formula

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6.3 Central Limit Theorem

Sampling distribution of sample means– Distribution using the means computed from all

possible random samples of a specific size taken from a populations

Sampling error– The difference between the sample measure and

the corresponding population measures due to the fact that the sample is not a perfect representation of the population.

Properties of the Distribution of sample means

1. The mean of the sample means will be the same as the populations mean.

2. The standard deviation of the sample means will be smaller than the standard deviation of the population, and will be equal to the populations standard deviation divided by the square root of the sample size.

The Central limit Theorem

As the sample size n increases without limit, the shape of the distribution of the sample means taken with replacement from a population with a mean µ and the standard deviation σ will approach a normal distribution.

Formulas

Sample mean

Xz

/

Xz

n

Example

The average number of pounds of meat that a person consumes per year is 218.4 pounds. Assume that the standard deviation is 25 pounds and the distribution is approximately normal.– Find the probability that a person selected at random

consumes less than 224 pound per year.– If a sample of 40 individuals is selected, find the

probability that the mean of the sample will be less than 224 pounds per year.

No given sample or under 30 TI-83

Left– Normalcdf(-E99,score,µ,σ)

Right– Normalcdf(score, E99,µ,σ)

Between 2 scores– Normalcdf(little score, big score,µ,σ)

Given sample 30 + TI-83

Left– Normalcdf(-E99,score,µ,(σ/ ))

Right– Normalcdf(score, E99,µ,(σ/ ))

Between 2 scores– Normalcdf(little score, big score,µ,(σ/ ))

n

n

n

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#’s 8-13

Normal Approximation to the Binomial Distribution

Binomial Normal (used for finding X)

P(X = a) P(a – 0.5 < X < a + 0.5)

P(X ≥ a) P(X > a – 0.5)

P(X > a) P(X > a + 0.5)

P(X ≤ a) P(X < a + 0.5)

P(X < a) P(X < a – 0.5)

Requirement: n*p ≥ 5 and n*q ≥ 5

xz

Practice

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