Normal Distributions

Section 4.4

Normal Distribution FunctionAmong 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 andsymmetric, 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 DistributionIf 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 thepressure 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 gaugesrejected.

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 forP(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 thata 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 + σP(μ – σ X μ + σ) =P(–1 Z 1) = .6826

μ – 2σ μ μ + 2σP(μ – 2σ X μ + 2σ) =P(–2 Z 2) = .9544

μ – 3σ μ μ + 3σP(μ – 3σ X μ + 3σ) =P(–3 Z 3) = .9974