Chapter 3: Probability Distributions and Statistics
Section 3.4
3.4 Normal Distribution
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Probability
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So far, we have focused on discrete random variables. Now we turn our attention to
continuous random variables. Continuous random variables cannot be represented using
histograms or tables, so we have to find another way to understand continuous random
variables.
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A
, or PDF,
of a continuous variable is a function that describes the relative likelihood for this random
variable to take on a given value.
Characteristics of PDFs:
• The PDF is always non-negative (either positive or zero).
• P (a ≤ X ≤ b) = area under the probability density function’s curve between the
values x = a and x = b.
• P (X = k) = 0. (Notice that there are infinitely many points to spread the probability over.)
• The total area under the curve is 1.
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A random variable X has a
distribution with mean µ and
standard deviation σ if the probability P (a ≤ X ≤ b) that X is between a and b is the
area under the PDF given by
(x−µ)2
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y = √ e− 2σ2
σ 2π
Probability
Note: We call this function a normal distribution curve. Here is it’s graph:
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Characteristics of the normal distribution curve:
1. It is bell shaped.
2. The curve is symmetric about x = µ and has a maximum there.
3. It lies above the x-axis.
4. It approaches but is never equal to 0 on both the positive and negative x-axis.
5. The area under the curve is exactly one.
6. The standard deviation determines how “steep” or “flat” the curve is (the curve is
flatter for large σ).
Why we care about the normal distribution: Many naturally occurring phenomena
obey a normal distrubution (or close to it). Some examples are:
• The weights of house cats
• The heights of women in Spain
• The waiting time in a line at a bank
• The diameter of a ball bearing coming off an assembly line
• The pounds of fertilizer in a 100-pound bag
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A
normal distribution is a normal distribution with
µ = 0 and σ = 1. We usually denote the random variable by Z.
It is very difficult to find the area under a normal curve. Therefore, we must use our
calculator to do it.
Finding Probability of Normal Distributions Using Your Calculator:
(1) Press DISTR (2nd → VARS).
(2) Select normalcdf( (Option 2).
(3) Enter the lower bound of the interval, comma key, upper bound of the interval,
comma, the mean, comma, the standard deviation, and close the parentheses.
(4) Press Enter.
Example 1 Let Z be a random variable that has the standard normal distribution
a) Shade the region whose area is P (X ≤ 0).
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b) Shade the region whose area is P (−2 ≤ X ≤ 2).
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c) Shade the region whose area is P (−3 ≤ X ≤ 1).
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Example 2 Let Z be a random variable that has the standard normal distribution. Find
the following. (Round to 4 decimal places)
a) P (Z < 0)
b) P (−1 ≤ Z ≤ 1)
c) P (−2.31 ≤ Z ≤ 2.31)
d) P (0.75 < Z ≤ 1.17)
e) P (Z < 2.31)
f) P (Z ≥ 2.31)
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Example 3 Assume X is distributed normally with mean µ = 70 and standard deviation
σ = 5.25. Find the following. (Round to 4 decimal places)
a) P (X < 70)
b) P (64.75 < X < 75.25)
c) P (35 < X < 88)
d) P (X < 86.54)
e) P (X > 72.30)
Sometimes we want to solve the opposite problem. We know the probability (or area)
that we want but don’t know the interval.
Finding the Value a so that P (X ≤ a) = k:
First, a picture:
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(1) Press DISTR (2nd → VARS).
(2) Select invNorm( (Option 3).
(3) Enter the area below the value a (i.e. the known value), comma, the mean, comma,
the standard deviation, and close the parentheses.
(4) Press Enter.
Example 4 Let Z be a random variable that has the standard normal distribution.
a) Find a so that P (Z ≤ a) = 0.6754. (Round to 4 decimal places)
b) Find a so that P (Z ≥ a) = 0.6754. (Round to 4 decimal places)
c) Find a so that P (−a ≤ Z ≤ a) = 0.6754. (Round to 4 decimal places)
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Example 5 Assume X is distributed normally with mean µ = 70 and standard deviation
σ = 5.25.
a) Find a so that P (X ≤ a) = 0.6754. (Round to 4 decimal places)
b) Find a so that P (X ≥ a) = 0.6754. (Round to 4 decimal places)
c) Find a and b symmetric about the mean so that P (−a ≤ X ≤ b) = 0.6754. (Round to
4 decimal places)
Example 6 According to data for a certain city, the weekly earnings of workers are
normally distributed with a mean of $700 and a standard deviation of $60.
a) What is the probability that a worker selected at random from the city makes more
than $805?
b) What minimum weekly earnings would put you in the top 15% of wage earners? Round
your answer to the nearest cent.
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c) What symmetric interval of wages about the mean comprises 64% of wage earners?
Round your answer to the nearest cent.
Example 7 The amount of soda in a 16-ounce can is normally distributed with a mean
of 16 ounces and a standard deviation of 0.50 ounces. What percentage of these cans will
have less than 15 ounces? Round your answer to 2 decimal places.
Example 8 The exam grades on a recent Math 166 exam were distributed normally with
a mean of 68 and a standard deviation of 15. What percentage of students should have
scored
a) a 90 or higher? Round your answer to 2 decimal places.
b) less than a 60? Round your answer to 2 decimal places.
c) between a 65 and 85, inclusive? Round your answer to 2 decimal places.
d) What score would a student need to have made on the exam to be in the top 10% of
the class? Round your answer to 2 decimal places.
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