Section 7.6: Normal Distributions

Standard normal distribution is the normal distribution with μ = 0 and σ = 1. The corresponding density curve is called the standard normal curve. It is customary to use the letter z to represent a variable whose distribution is described by the standard normal curve. The term z curve is often used in place of standard normal curve.

Using the table of standard normal curve areas For any number z* between -3.89 and 3.89 and rounded to two decimal places, Appendix Table 2 gives (area under z curve to the left of z*) = P(z < z*) = P(z ≤ z*) Where the letter z is used to represent a random variable whose distribution is the standard normal distribution

To find this probability, locate the following: The row labeled with the sign of z* and the digit to either side of the decimal point (e.g., -1.7 or 0.5) The column identified with the second digit to the right of the decimal point in z* (e.g., .06 if z* = -1.76) The number at the intersection of this row and column is the desired probability, P(z < z*).

Using the Normal Tables Find P(z < 0.46) Column labeled 0.06 Row labeled 0.4 P(z < 0.46) = 0.6772

Sample Calculations Using the Standard Normal Distribution Using the standard normal tables, find the proportion of observations (z values) from a standard normal distribution that satisfy each of the following: P(z < 1.83) = 0.9664 (b) P(z > 1.83) = 1 – P(z < 1.83) = 1 – 0.9664 = 0.0336

Finding Probabilities To calculate probabilities for any normal distribution, standardize the relevant values and then use the table of z curve areas. More specifically, if x is a variable whose behavior is described by a normal distribution with mean μ and standard deviation σ, then

Example In poor countries, the growth of children can be an important indicator of general levels of nutrition and health. Data suggest that a reasonable model for the probability distribution of the continuous numerical variable x = height of a randomly selected 5-year old child is a normal distribution with a mean of μ = 100 cm and standard deviation σ = 6 cm. What proportion of the heights is between 94 and 112 cm?

Example The Environmental Protection Agency has developed a testing program to monitor vehicle emission levels of several pollutants. An article describes the program, which measures pollutants from various types of vehicles, all following a fixed driving schedule. Data from the article suggest that the emissions of nitrogen oxides, which are major factors of smog, can be plausibly modeled using a normal distribution. Let x denote the amount of this pollutant emitted by a randomly selected vehicle. The distribution of x can be described with μ = 1.6 and σ = .4.

Suppose that the EPA wants to offer some sort of incentive to get the worst polluters off the road. What emission levels constitute the worst 10% of the vehicles? The worst 10% would be the 10% with the highest emissions level.

For the standard normal distribution, the largest 10% are those with z values greater than z* = 1.28 (From Appendix Table 2, based on a cumulative area of .90). Then