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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