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```The Normal Distribution
Normal and Skewed
Distributions
Normal Distribution
• Definition: A continuous, symmetric, bell-shaped
distribution of a variable.
• The shape and position of the curve depend on 2
variables. ( The mean and the standard deviation)
• The larger the deviation the more dispersed, or
• The area under the curve is more important than
the frequencies themselves. When pictured the
y-axis is usually omitted.
• Z-value: The number of standard deviations that
a particular X value is away from the mean.
Other Properties of
Normal Distribution
• 1. The normal distribution curve is unimodal.
• 2. Theoretically, no matter how far it extends, it
never touches the x-axis. It just gets increasingly
closer.
• 3. The total area, under the curve, is equal to 1.00
or 100%.
• 4. The area under the curve follows the Empirical
Rule. One deviation about 68% of the curves area.
Two deviations about 95% of curves area. Finally
3 deviations about 99.7% of curves area.
Standard Normal Distribution
• Differs slightly from normal distribution in the
fact that its’ mean value is 0 and the standard
deviation value is 1.
Finding Areas
Under the Curve
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There are 7 basic types of problems.
1. Between 0 and any z-value.
2. In any tail of the curve.
3. Between two z-values on the same side of the
mean.
4. Between z-values on opposite sides of the
mean.
5. Less than any z-value to the right of the mean.
6. Greater than any z-value to the left of the mean
7. In any two tails of the curve.
How to Solve for the
Area
• Step 1: Draw the picture of the curve.
• Step 2: Shade the desired area.
• Step 3: Depending on the area shaded, you
may either do nothing, add, or even subtract
appropriate z-values from Table E.
Examples
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#1: Find the area between z = 0 and z = - 2.4
#2: Find the area to the left of z = 1.25
#3: Find the area between z = .85 and z = 1.7
#4: Find the area between z = - .54 and z = 2
#5: Find the area to the left of z = - 2.07 and
to the right of z = 3.04
Probability Questions
• Area under the curve corresponds to a
probability.
• Problems involving probabilities are solved in
the same manner as previous area problems.
• Examples:
• P( 0 < n < 2.34 )
• P ( n > - 1.78 )
• P ( n < .85 )
Find the Z
• May need to find the specific z-value for a given area
under the normal distribution.
• Work backwards:
• Step 1: Find the area in Table E.
• Step 2: Read the correct z-value in the left column and
in the top row and add two values together.
• Example 1: Find the z-value that corresponds to the
area 0.4931.
• Example 2: Find the z-value that corresponds to the
area 0.2794.
Other Applications
• The standard normal distribution curve can be used to
solve a wide variety of problems. Including
probabilities and finding specific data values for given
percentages. Only requirement is that the variable be
normally or approximately distributed.
• To solve problems we need to transform the original
variable into a standard normal distribution variable
and then use Table E to solve the problems.
• To determine z-value we use the same formula from
Chapter 3. z = ( value – mean ) ÷ sd.
Examples
• #1 The avg. hourly wage of Detroit car workers is
\$13.50. If standard deviation is \$1.80 find the
probabilities for a randomly selected worker. Assume
variable is normally distributed.
• P(The worker earns more than \$15.60)
• P(The worker earns less than \$9.00)
• #2 The avg. age of CEOs is 62 years. If standard
deviation is 3 years find the probability for randomly
selected CEOs will fall in the following ranges.
Assume variable is normally distributed.
• Between 64 and 70 years old.
• Between 57 and 62 years old.
Steps to finding
specific data values
• Work backwards.
• Step 1: Find the area in Table E.
• Step 2: Read the correct z-value in the left column and
in the top row and add two values together.
• Step 3: Use the formula X = z ● 𝜎 + 𝜇
• 𝜎 = 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
• 𝜇 = 𝑀𝑒𝑎𝑛
Examples
• The scores on a test has a mean of 90 and
standard deviation of 7. If a teacher wishes to
select the top 30% of students who took the test
find the cutoff score. Assume variable is
normally distributed.
• A pet shop owner decides to sell snakes that
appeal to the middle 50% of customers. The
mean for the snakes is \$37.90 and the standard
deviation is \$5.60. Find the minimum and
maximum prices of the snakes the owner should
sell. Assume variable is normally distributed.
Distribution of Sample Means
• Along with knowing how data values vary about a mean for
a population, statisticians are also interested in knowing
about the distribution of the means of samples taken.
• Sampling distribution of sample means: A distribution
obtained by using the means computed from random
samples of a specific size taken from a population.
• If randomly selected, the sample means, for most part are
somewhat different from the population mean. These
differences are caused by sampling error.
• Sampling Error: The difference between the sample
measure and the corresponding population measure due to
the fact sample is not a perfect representation of the
population
Properties of the Distribution of
Sample Means
• If all possible samples of a specific size are selected
from a population, the distribution of the sample means
has 2 important properties.
• #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. It
will be equal to the population standard deviation
divided by the square root of the sample size.
• Standard deviation of the sample means is called the
standard error of the mean.
Central Limit Theorem
• A third property of the sampling distribution of
sample means pertains to the shape of the
distribution.
• CLT definition: As sample size increases, the
shape of the distribution of the sample means
taken from a population with mean 𝜇 and standard
deviation 𝜎 approach a normal distribution.
means in the same way that normal distributions
• Major difference: z = ( Value - 𝜇 ) ÷ ( 𝜎 ÷ √ n )
Examples
• The avg. age of teachers is 46 years, with a standard
deviation of 6 years. Find the probability, that the mean
of the sample of 7 randomly selected teachers is greater
than 50 years old. Assume the variable is normally
distributed.
• The mean weight of 13 year old males is 108 pounds
and the standard deviation is 11. If a sample of 35
males is selected find the probability that the mean of
the sample will be less than 104 pounds. Assume the
variable is normally distributed.
Examples
• A survey found that Americans generate an average of
17.2 pounds of glass garbage each year. The standard
deviation is 2.5 pounds. Find the probability that the
mean of a sample of 55 families will be between 17 and
18 pounds.
• Average rainfall for Des Moines is 30.83 inches with a
standard deviation of 5 inches. If a random sample of
10 years is selected find the probability that the mean
will be between 32 and 33 inches.
• Mean score on a dexterity test for 12-year olds is 30.
Standard deviation is 5. If test administered to 22
students find probability mean of samples will be
between 27-31.
The Correction Factor
• The formula for standard error of the mean,
𝜎 ÷ √n, is accurate for samples drawn with/without
replacement from very large populations.
• The correction factor is needed if relatively large
samples are taken from a small population.
• The sample mean will more accurately estimate the
population mean and there will be less error in the
estimation.
• Generally used when the sample is greater than 5% of
the population.
The Correction Factor
Formulas
• Formula for correction factor:
𝑁 −𝑛
•
𝑁 −1
• Formula for computing the z-value:
Value - 𝜇
𝑁 −𝑛
(𝜎 ÷ √n ) ●
𝑁 −1
Examples
• #1: A study of 100 renters showed that the
average rent of their apartments was \$1,200 and
the standard deviation was \$250. If 28 renters are
selected, find the probability the average rent of
their apartments was greater than \$1,300.
• #2: The average price of a new suit at Macy’s is
\$275 and a standard deviation of \$10. If 17 suits
are sold from a lot of 70, find the probability that
the average price will be less than \$270?
The Normal Approximation to
the Binomial Distribution
• The normal distribution is used to solve problems that
involve the binomial distribution due to the fact that
when n is rather large the calculations get to difficult to
do by hand.
• When p is close to 0.5 and as n increases the binomial
distribution becomes similar to the normal distribution.
• When p is close to 0 or 1 and n is smaller the normal
approximation becomes inaccurate.
• Due to the fact above the normal approximation should
only be used when n ● p and n ● q are both greater than
or equal to 5.
Formulas for Mean and Standard
Deviation of Binomial Distribution
• Mean: 𝜇 = 𝑛 ● 𝑝
• Standard Deviation: 𝑛 ● 𝑝 ● 𝑞
Steps for Normal Approximation to the
Binomial Distribution
• Step 1: Check to see whether normal approximation
can be used.
• Step 2: Find the mean and standard deviation.
• Step 3: Write the problem in probability notation.
Remember to use the continuity correction factor, and
show corresponding area under the curve.
• Step 5: Find the corresponding z-values.
• Step 6: Find solution.
• Correction factor: Employed when a continuous
distribution is used to approximate a discrete
distribution. Works like lower and upper boundaries.
Examples
• Of all 5 to 6 year-old children, in NJ, 87% are
enrolled in school. If a sample of 300 such
children are randomly selected, find the
probability that at least 247 will be enrolled in
school.
• A survey found that 17% of teenage drivers
text while driving. If 400 drivers are selected
at random, find the probability that exactly 65
say they text while driving.
Examples
• Use the normal approximation to the binomial
to find the probabilities for specific value/s of
X.
• #1: n = 25, p = .65, X = 12
• #2: n = 80, p = .4, X = 34
• #3: n = 30, p = .5 X ≤ 8
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