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The Normal Distribution
Business Statistics
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• Probability density functions
• The Standard Normal Distribution
• The z-scores
• The Empirical Rule
• Tables of the Standard Normal Distribution
• Examples and more examples
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Probability Density Functions
• A PDF, or density, describes the relative
likelihood for a random variable to take on a
given value.
• The probability of the random variable falling
within a particular range of values is given by
the area under the density function but
above the horizontal axis and between the
lowest and greatest values of the range.
Probability Density Functions
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The Normal Distribution
http://statsthewayilikeit.com/
The Normal Distribution
• The normal distribution is the most
important of all the distributions, and it
occurs naturally in many instances, in nature
and sciences, in particular, in statistics.
• One of the characteristic features of a
normal distribution is the bell-shaped curve.
• Each normal distribution has two
parameters: the mean 𝜇 (“mu”) and the
standard deviation 𝜎 (“sigma”). We denote
such a distribution by 𝑁(𝜇, 𝜎).
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The Standard Normal Distribution
• The standard normal distribution 𝑁(0,1) has
the mean zero and the standard deviation 1.
The Empirical Rule
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The z-score
• Any normal distribution can be converted to
the standard normal distribution by means of
the z-score.
• The formula:
𝑧=
𝑥−𝜇
.
𝜎
• If the z-score is positive, it means that the
value of x is bigger than the mean 𝜇. On the
graph, it to the right of 𝜇 . And if it is negative,
then x is smaller, or to the left of 𝜇.
The z-score
• The z-score allows us to compare data that
are scaled differently.
• The value of the z-score tells you how many
standard deviations the value x is above (to
the right of) or below (to the left of) the
mean, μ.
• Example. Suppose that 𝑋 ~ 𝑁(14, 5) .
• (What does this notation mean?)
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Example (continued)
• 𝑋 ~ 𝑁(14, 5)
• Suppose that 𝑥 = 22, its z-score
22 − 14
= 1.6
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• It means that the value of x is 1.6 times the
standard deviation units to the right of the
mean.
• For x = 4, the z-score z = −2. It means that …
𝑧=
The z-score
• If we know the z-score, we can figure out the
value of x: 𝑥 = 𝜇 + 𝑧 ∙ 𝜎.
• Example. If 𝑋 ~ 𝑁(64, 12.5), between what
two values lie the values of about 68% of the
values of x ? Answer: between 51.5 and 76.5
• What about 95% of the values of x ?
Answer: between 39 and 89 .
• What about 99.7% of the values of x ?
Answer: between 26.5 and 101.5 .
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Using the normal distribution
The area to the right is complimentary to the area to the left:
𝑃 𝑋 > 𝑥 = 1 − 𝑃(𝑋 < 𝑥)
Tables of the Standard Normal Distribution
• For example:
http://www.mathsisfun.com/data/standard-normal-distribution-table.html
• How to use a table like this ?
1. Compute the z-score(s).
2. Find the table value(s) T(z) .
3. Decide, what is being asked:
T(z) , 0.5 − T(z), 0.5 + T(z),
T(z1) − T(z2), T(z1) + T(z2), …
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Using the table
Using the table
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Using the table
Same goes for negative z’s by symmetry !
Using the table
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Using the table
Examples
• What is the area between z = 0 and z = 1.43 ?
𝑃 0 < 𝑧 < 1.43 = 𝑇 1.43 = 0.4236
• What is the area to the right of z = 0.28 ?
𝑃 𝑧 > 0.28 = 0.5 − 𝑇 0.28
= 0.5 − 0.1103 = 0.3897
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Examples
• What is the probability that z < ̶ 0.98 ?
𝑃 𝑧 < −0.98 = 0.5 − 𝑇 0.98
= 0.5 − 0.3365 = 0.1635 or 16.35%
• Find the probability that −0.4 < 𝑧 < 0.65 .
𝑃 −0.4 < 𝑧 < 0.65 = 𝑇 0.65 + 𝑇 0.4 =
0.2422 + 0.1554 = 0.3976 or 39.76%
Examples
• What is the area between 𝑧 = −2.03 and
𝑧 = −1.17 ?
𝑃 −2.03 < 𝑧 < −1.17 = 𝑇 2.03 − 𝑇(1.17)
= 0.4788 − 0.3790 = 0.0998
• What is the area to the right of 𝑧 = −0.66 ?
𝑃 𝑧 > −0.66 = 0.5 + 𝑇 0.66
= 0.5 + 0.2454 = 0.7454
etc.
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Example
• The weights of adult indri lemurs are
normally distributed with a mean of 13.8 lbs
and a standard deviation of 2.2 lbs.
Given :
𝜇 = 13.8 lbs
𝜎 = 2.2 lbs
Example (continued)
• What is the probability that a randomly
selected adult indri weighs less than 15 lbs?
1. Compute the z-score: 𝑧 =
15−13.8
2.2
= 0.55
2. Draw a bell-shaped curve and indicate the
region.
3. Compute the answer:
𝑃 𝑧 < 0.55 = 0.5 + 𝑇 0.55 =
0.5 + 0.2088 = 0.7088
or 70.88%
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Example (continued)
• What percentage of adult indris weigh more
than 18.5 lbs.?
1. Compute 𝑧 =
18.5−13.8
2.2
= 2.14
2. Draw a bell-shaped curve + region.
3. Compute the answer:
𝑃 𝑧 > 2.14 = 0.5 − 𝑇 2.14 =
0.5 − 0.4838 = 0.0162 or 1.62%.
Example (continued)
• Try it yourself !
• What percentage of adult indris weigh more
than 12.5 lbs.?
Answer: 72.24%
• What is the probability that a randomly
selected adult indri weighs less than 13 lbs.?
Answer: 35.94%
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Example (continued)
What percentage of adult indris weigh
between 11.4 and 15.8 lbs.?
1. Compute 𝑧1 = 0.91 and 𝑧2 = −1.09
2. Graph the curve and the region.
3. Answer: 𝑃 −1.09 < 𝑧 < 0.91 =
𝑇 0.91 + 𝑇 1.09 = 0.6807 or 68.07%
Try it: what is the probability that an adult
indri weighs between 12 and 13 lbs.?
Answer: 15.33%
Finding the percentiles.
• Find the percentile for a data value is about
the same as finding the probability that a
random data value is below the given one,
and rounding off to the nearest percentage.
• For example, if the weight of an adult indri is
15 lbs., then he/she is in the 71st percentile.
• Try it: if an adult indri weighs 11.7 lbs., to
which percentile does she or he belong?
Answer: he or she belongs to the 17th percentile.
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Reading the table backwards
• Suppose that we want to find a data value
corresponding to the kth percentile.
• 𝑇(𝑧) is either
𝑘−50
100
or
50−𝑘
100
(must be > 0).
• Find the value inside the table closest to 𝑇(𝑧),
and find the corresponding z-score.
(If there are two closest, take the average.)
• Finally, use the formula 𝑥 = 𝜇 + 𝑧 ∙ 𝜎 to find
the data value x.
Example
• Find the weight of an adult indri in the 28th percentile.
• We have:
50−28
100
= 0.22 . In the table, the value
most close to 0.22 is 0.2190, corresponding to 𝑧 = 0.58.
• But we must take the negative value 𝑧 = −0.58 ,
because the data value is to the left of the mean!
• Finally, 𝑥 = 13.8 + −0.58 ∙ 2.2 = 12.52 lbs.
• Try it: the top 33% of adult indris weigh at least how
much?
Answer: 14.77 lbs.
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