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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 #
Page 311-312 #’s
10-13, 20-23, 30-33, 38-39
Finding the area under the Curve

Between two z scores
–

Tails of two z scores
–

Bigger z chart # - smaller z chart #
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:
–

To the right
–

1- given percent, then use chart
Between two z’s
–

Find percent in chart then find z score
.5- percent/2, then chart
Tails:
–
Percent/2, then chart
Page 312-313 #’s 46 - 49
Using TI-83 Plus

To the left:
–

To the right
–

invNorm(1- given percent)
Between two z’s
–

invNorm(percent)
invNorm(.5- percent/2)
Tails
–
invNorm(percent/2)
invNorm(
1.
2.
3.
4.
Hit 2nd Button
Hit DISTR
Hit 3 key or arrow down to invNorm
Type in formula
Page 312-313 #’s 46 - 49
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.
2.
The mean of the sample means will be the
same as the populations mean.
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
z

X 

Sample mean
X 
z
/ 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
–

Right
–

Normalcdf(-E99,score,µ,σ)
Normalcdf(score, E99,µ,σ)
Between 2 scores
–
Normalcdf(little score, big score,µ,σ)
Given sample 30 +

Left
–

Normalcdf(-E99,score,µ,(σ/ n))
Right
–

TI-83
Normalcdf(score, E99,µ,(σ/ n ))
Between 2 scores
–
Normalcdf(little score, big score,µ,(σ/ n ))
Page 338-339

#’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
z
x

Practice
Page 346-347 2-3