Normal Distributions
Section 4.4
Normal Distribution Function
Among all the possible probability density functions, there is an important
class of functions called normal density functions, or normal distributions.
The graph of a normal density function is bell-shaped and
symmetric, as the following figure shows.
Normal Distribution Function
• A normal distribution, is a function of the form
• where μ is the mean and σ is the standard deviation.
Graphs of Normal Distributions
•
The “inflection points” are the points where the curve changes from bending in one direction
to bending in another.
•
Figure below shows the graph of several normal distribution functions.
•
The third of these has mean 0 and standard deviation 1, and is called the standard normal
distribution.
•
We use Z rather than X to refer to the standard normal variable.
Calculate Probability using Standard
Normal Distribution function
• The standard normal distribution has μ = 0 and σ = 1.
The corresponding variable is called the standard
normal variable, which we always denote by Z.
• Recall that to calculate the probability P(a ≤ Z ≤ b), we
need to find the area under the distribution curve
between the vertical lines z = a and z = b.
• We can use the table in the Appendix to look up these
areas. Here is an example.
Example 1
• Let Z be the standard normal variable. Calculate
the following probabilities:
• a. P(0 ≤ Z ≤ 2.4)
• b. P(0 ≤ Z ≤ 2.43)
• c. P(–1.37 ≤ Z ≤ 2.43)
• d. P(1.37 ≤ Z ≤ 2.43)
Example 1a
• We are asking for the shaded area under the standard
normal curve shown in the graph below.
• We can find this area, correct to four decimal places, by
looking at the table in the Appendix, which lists the area
under the standard normal curve from Z = 0 to Z = b for any
value of b between 0 and 3.09.
Example 1a continued
• To use the table, write 2.4 as 2.40, and read the entry
in the row labeled 2.4 and the column labeled
.00 (2.4 + .00 = 2.40). Here is the relevant portion of
the table:
• Thus, P(0 ≤ Z ≤ 2.40) = .4918.
Example 1b
• The area we require can be read from the same portion
of the table shown above. Write 2.43 as 2.4 + .03, and
read the entry in the row labeled 2.4 and the column
labeled .03:
• Thus, P(0 ≤ Z ≤ 2.43) = .4925.
Example 1c
• Here we cannot use the table directly because
the range –1.37 ≤ Z ≤ 2.43 does not start at 0.
But we can break the area up into two smaller
areas that start or end at 0:
•
P(–1.37 ≤ Z ≤ 2.43) = P(–1.37 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 2.43).
• In terms of the graph, we are
splitting the desired area into
two smaller areas.
Example 1c continued
• We already calculated the area of the right-hand piece in part (b):
•
P(0 ≤ Z ≤ 2.43) = .4925.
• For the left-hand piece, the symmetry of the normal curve tells us that
•
P(–1.37 ≤ Z ≤ 0) = P(0 ≤ Z ≤ 1.37).
• This we can find on the table. Look at the row labeled 1.3 and the column
labeled .07, and read
•
P(–1.37 ≤ Z ≤ 0) = P(0 ≤ Z ≤ 1.37)
•
•
= .4147.
• Thus,
•
P(–1.37 ≤ Z ≤ 2.43) = P(–1.37 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 2.43)
•
= .4147 + .4925
•
= .9072.
Example 1d
• The range 1.37 ≤ Z ≤ 2.43 does not contain 0,
so we cannot use the technique of part (c).
Instead, the corresponding area can be
computed as the difference of two areas:
• P(1.37 ≤ Z ≤ 2.43) = P(0 ≤ Z ≤ 2.43) – P(0 ≤ Z ≤ 1.37)
•
= .4925 – .4147
•
= .0778.
Calculating probability of any normal
distribution
• Although we have tables to compute the area under the standard normal
curve, there are no readily available tables for nonstandard distributions.
• For example, If μ = 2 and σ = 3, then how would we calculate P(0.5 ≤ X ≤
3.2)? The following conversion formula provides a method for doing so:
• Standardizing a Normal Distribution
If X has a normal distribution with mean μ and standard deviation σ , and
if Z is the standard normal variable, then
Quick example
• If μ = 2 and σ = 3, then
•
= P(−0.5 ≤ Z ≤ 0.4)
•
= .1915 + .1554
•
= .3469.
Example 2
• Pressure gauges manufactured by Precision Corp. must be
checked for accuracy before being placed on the market.
• To test a pressure gauge, a worker uses it to measure the
pressure of a sample of compressed air known to be at a
pressure of exactly 50 pounds per square inch. If the gauge
reading is off by more than 1% (0.5 pounds), it is rejected.
Assuming that the reading of a pressure gauge under these
circumstances is a normal random variable with mean 50 and
standard deviation 0.4, find the percentage of gauges
rejected.
Example 2 solution
• If X is the reading of the gauge, then X has a normal
distribution with μ = 50 and σ = 0.4. We are asking for
P(X < 49.5 or X > 50.5) = 1 – P(49.5 ≤ X ≤ 50.5).
• We calculate
•
•
= P(–1.25 ≤ Z ≤ 1.25)
•
= 2  P(0 ≤ Z ≤ 1.25)
•
= 2(.3944)
•
= .7888.
Example 2 solution continued
• So, P(X < 49.5 or X > 50.5) = 1 – P(49.5 ≤ X ≤ 50.5)
•
•
•
= 1 – .7888
•
= .2112.
• In other words, about 21% of the gauges will be
rejected.
Calculating probability with any
normal distribution
• In many applications, we need to know the probability that
a value of a normal random variable will lie within one
standard deviation of the mean, or within two standard
deviations, or within some number of standard deviations.
• To compute these probabilities, we first notice that, if X has
a normal distribution with mean μ and standard deviation
σ, then
•
P(μ – kσ ≤ X ≤ μ + kσ) = P(–k ≤ Z ≤ k)
• by the standardizing formula.
1, 2 and 3 standard deviations from
the mean.
• We can compute these probabilities for
various values of k using the table in the
Appendix, and we obtain the following results.
• Probability of a Normal Distribution Being
within k Standard Deviations of Its Mean
μ–σ
μ μ+σ
P(μ – σ  X  μ + σ) =
P(–1  Z  1) = .6826
μ – 2σ
μ
μ + 2σ
P(μ – 2σ  X  μ + 2σ) =
P(–2  Z  2) = .9544
μ – 3σ μ μ + 3σ
P(μ – 3σ  X  μ + 3σ) =
P(–3  Z  3) = .9974