THE NORMAL
DISTRIBUTION
CHAPTER 6
INTRODUCTION
• The normal distribution is used often by researchers to
determine normal intervals for specific medical tests
• Many continuous variables have distributions that are bellshaped
• This type of distribution is known as a bell curve or a
Gaussian distribution
(named for German mathematician Carl Friedrich Gauss)
KEY TERMS
• Symmetric distribution
• When the data values are evenly distributed about the
mean
• Negatively or left-skewed distribution
• Majority of data values fall to the right of the mean, mean
is left of the median, mean and median are left of the mode
• Positively or right-skewed distribution
• Majority of data values fall to the left of the mean, mean is
right of the median, mean and median are right of mode
6.1 – NORMAL DISTRIBUTIONS
• A theoretical curve, called a normal distribution can be
used to study many variables that are not perfectly
normally distributed but are approximately normal
• Mathematical expression for a normal distribution
𝒚=
𝑿−𝝁 𝟐
−
𝒆 𝟐𝝈𝟐
𝝈 𝟐𝝅
• Where
𝒆 ≈ 𝟐. 𝟕𝟏𝟖
𝝁 = 𝒑𝒐𝒑𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒎𝒆𝒂𝒏
𝝅 ≈ 𝟑. 𝟏𝟒
𝝈 = 𝒑𝒐𝒑𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
NORMAL DISTRIBUTION
• Normal distribution
• A continuous, symmetric, bell-shaped distribution of a
variable
• Shape and position of a normal distribution curve depend
on two parameters, the mean and the standard deviation
PROPERTIES OF NORMAL
DISTRIBUTIONS
1.
Normal distribution curve is bell-shaped
2.
Mean, median, and mode are equal are located at center of
the distribution
3.
Normal distribution curve is unimodal
4.
Curve is symmetric about the mean
5.
Curve is continuous, no gaps or holes
6.
Curve never touches the X-axis
7.
Total area under a normal distribution curve is 1.00 or 100%
8.
Area under the part of a normal curve that lies within 1
standard deviation of the mean is approx. 68%, within 2
standard deviations approx. 95%, and within 3 standard
deviations, approx. 99.7%
STANDARD NORMAL
DISTRIBUTION
• Standard normal distribution
• A normal distribution with a mean of 0 and a standard
deviation of 1
• All normally distributed variables can be transformed into
the standard normally distributed variable by using the
standard score formula:
𝒛=
𝒗𝒂𝒍𝒖𝒆−𝒎𝒆𝒂𝒏
𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
=
𝑿−𝝁
𝝈
PROCEDURE FOR FINDING AREA
UNDER NORMAL CURVE
• The z value is actually the number of standard deviations
that a particular X value is away from the mean. Table E in
Appendix C gives the area under the curve for any z value
Procedure:
1. To the left of any z value:
1. Look up the z value in the table and use the area given
2. To the right of any z value:
1. Look up the z value and subtract the area from 1
3. Between any two z values:
1. Look up both z values and subtract the corresponding
areas
EXAMPLES
• 6–1
• Find the area to the left of z = 1.99
• 6–2
• Find the area to the right of z = -1.16
• 6–3
• Find the area between z = +1.68 and z = -1.37
EXAMPLES
•
A normal distribution curve can be used as a probability
distribution curve for normally distributed variables
•
•
6–4
•
•
•
•
NOTE: In a continuous distribution, the probability of any
exact z value is 0
A. 𝑃 0 < 𝑧 < 2.32
B. 𝑃 𝑧 < 1.65
C. 𝑃 𝑧 > 1.91
6–5
•
Find the z value such that the area under the standard normal
distribution curve between 0 and the z value is 0.2123
6.2 – APPLICATIONS OF THE
NORMAL DISTRIBUTION
• To solve problems by using the standard normal
distribution, transform the original variable to a standard
normal distribution variable by using the z score formula
EXAMPLES
• 6–6
• A survey by the National Retail Federation found that
women spend on average $146.21 for the Christmas
holidays. Assume the standard deviation is $29.44. Find
the percentage of women who spend less than $160.00.
Assume the variable is normally distributed.
EXAMPLES
• 6–7
• Each month, an American household generates an
average of 28 pounds for newspaper for garbage or
recycling. Assume the standard deviation is 2 pounds. If a
household is selected at random, find the probability of its
generating
•
•
A. between 27 and 31 pounds per month
B. more than 30.2 pounds per month
EXAMPLE 6 – 9
• To qualify for a police academy, candidates must score in the
top 10% on a general abilities test. The test has a mean of 200
and a standard deviation of 20. Find the lowest possible score
to qualify. Assume the test scores are normally distributed.
EXAMPLES
• Formula for finding X
• When you must find the value of X, you can use the
following formula:
𝑿=𝒛∗𝝈+𝝁
•
6 – 10
•
For a medical study, a researcher wishes to select people
in the middle 60% of the population based on blood
pressure. If the mean systolic blood pressure is 120 and
the standard deviation is 8, find the upper and lower
reading that would qualify people to participate in the
study.
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 population
• Sampling error
• The difference between the sample measure and the
corresponding population measure due to the fact that the
sample is not a perfect representation of the population
PROPERTIES
• When all possible samples of a specific size are selected
with replacement from a population, the distribution of the
sample means for a variables has two important
properties:
Properties of the Distribution of Sample Means
1. The mean of the sample means will be the same as the
population mean
2. The standard deviation of the sample means will be
smaller than the standard deviation of the population,
and it will be equal to the population standard deviation
divided by the square root of the sample size
STANDARD ERROR OF
THE MEAN
• If all possible sample of size n are taken with replacement
from the same population, then
𝝁𝑿 = 𝝁
and the standard deviation of the sample means equals
𝝈𝑿 = 𝝈
𝒏
• Standard error of the mean
• Standard deviation of the sample means
CENTRAL LIMIT THEOREM
• The Central Limit Theorem
• As the sample size n increases without limit, the shape of
the distribution of the sample means take with replacement
from a population with mean μ and standard deviation σ
will approach a normal distribution. This distribution will
have a mean μ and a standard deviation 𝜎 𝑛
NEW FORMULA FOR
SAMPLE MEANS
• 𝒛=
𝑿−𝝁
𝝈
𝒏
• Two important things to remember when using the Central
Limit Theorem
1. When the original variable is normally distributed, the
distribution of the sample means will be normally
distributed, for any sample size n
2. When the distribution of the original variable might not be
normal, a sample size of 30 or more is needed to use a
normal distribution to approximate the distribution of the
sample means. The larger the sample, the better the
approximation will be
EXAMPLES
• 6 – 13
• A.C. Neilsen reported that children between the ages of 2
and 5 watch an average of 25 hours of television per week.
Assume the variables is normally distributed and the
standard deviation is 3 hours. If 20 children between the
ages of 2 and 5 are randomly selected, find the probability
that the mean of the number of hours they watch television
will be greater than 26.3 hours.
EXAMPLES
• 6 – 14
• The average age of a vehicle registered in the United
States is 8 years, or 96 months. Assume the standard
deviation is 16 months. If a random sample of 36 vehicles
is selected, find the probability that the mean of their age is
between 90 and 110 months.
EXAMPLES
• 6 – 15
• 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.
a. Find the probability that a person selected at random
consumes less than 224 pounds per year
b. 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
6.4 – NORMAL APPROXIMATION
TO THE BINOMIAL DISTRIBUTION
• A normal approximation to a binomial distribution is best
used when 𝒏 ∗ 𝒑 ≥ 𝟓 𝒂𝒏𝒅 𝒏 ∗ 𝒒 ≥ 𝟓
• In addition to this, a correction for continuity may be used
in the normal approximation
• Correction for continuity
• Correction employed when a continuous distribution is
used to approximate a discrete distribution
• For any specific value of X, the boundaries of X in the
binomial distribution must be used
SUMMARY OF NORMAL APPROXIMATION TO
BINOMIAL DISTRIBUTION
Binomial
Normal
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)
For all cases, 𝜇 = 𝑛 ∗ 𝑝, 𝜎 = 𝑛 ∗ 𝑝 ∗ 𝑞, 𝑛 ∗ 𝑝 ≥ 5, 𝑎𝑛𝑑 𝑛 ∗ 𝑞 ≥ 5
PROCEDURES
Procedure for Normal Approximation to Binomial Distribution
1. Check to see whether the normal approximation can be
used
2. Find the mean μ and the standard deviation σ
3. Write the problem in probability notation, using X
4. Rewrite the problem by using the continuity correction
factor, and show the corresponding area under the
normal distribution
5. Find the corresponding z values
6. Find the solution
EXAMPLES
• 6 – 16
• A magazine reported that 6% of American drivers read the
newspaper while driving. If 300 drivers are selected at
random, find the probability that exactly 25 say they read
the newspaper while driving.
EXAMPLES
• 6 – 17
• Of the members of a bowling league, 10% are widowed. If
200 bowling league members are selected at random, find
the probability that 10 or more will be widowed.
EXAMPLES
• 6 – 18
• If a baseball player’s batting average is 0.320 (32%), find
the probability that they players will get at most 26 hits in
100 times at bat.