Practice problems for these concepts can be found at:

### Standardizing a Normal Distribution

In order to find areas under a normal distribution having mean μ and standard deviation σ, the normal distribution must be standardized. A normal random variable X having mean μ and standard deviation σ is converted or transformed to a standard normal random variable by the formula given in (6.6).

**EXAMPLE 6.7** Figure 6-15 shows the distribution of adult male weights for a particular age group. The weights, X, are normally distributed with mean 170 pounds and standard deviation equal to 15 pounds. The standardized value for the weight For this age group, an individual who weighs 215 pounds is 3 standard deviations above average. The standardized value for 170 pounds is zero, and the standardized value for 125 pounds is –3.

### Applications Of The Normal Distribution

The fact that many real-world phenomena are normally distributed leads to numerous applications of the normal distribution. Applications of the normal distribution usually involve finding areas under a normal curve. *To find the area between two values of x for a normal distribution, first convert both values of x to their respective z values. Then find the area under the standard normal curve between those two z values. The area between the two z values gives the area between the corresponding x values.*

**EXAMPLE 6.8** In a study involving stress-induced blood pressure, volunteers played a computer game called the color-word interference task. The game was set so that everyone made errors about 17% of the time. The average increase in systolic blood pressure was 10 points of systolic pressure, and the standard deviation was 3 points. The percent experiencing an increase of 16 points or more is found by evaluating P(X > 16) and multiplying by 100. The probability is shown as the shaded area in Fig. 6-16. The z value corresponding to x = 16 is The area shown in Fig. 6-16 is the same as the area shown in Fig. 6-17.

The equality of the areas in Figures 6-16 and 6-17 is expressible in terms of probability as follows:

P(X > 16) = P(Z > 2)

Using the standard normal distribution table, we find that P(Z > 2) = .5 – .4772 = .0228. The percent experiencing an increase of 16 systolic points or more is 2.28%.

EXCEL may be used to find the area in Fig. 6-16. The built-in function =NORMDIST(16,10,3,1) entered into any Excel cell gives the area to the left of 16 in Fig. 6-16. Therefore, to find the area to the right of 16, use the fomula, =1-NORMDIST(16,10,3,1). The result is 0.02275. The first parameter in NORMDIST is the x value, the second is the mean of the normal distribution, the third is the standard deviation, and the fourth is 0 or 1. A 0 in the fourth position asks for the height up to the normal curve at x and a 1 asks for the area to the left of x. The result is P(X > 16) = 1 – P(X < 16) = 1 – 0.97725 or 0.02275. EXCEL also has the standard normal distribution function built in to the software. The standard normal distribution function is =NORMSDIST(z), and the area to the left of 16 in Fig. 6-16 is the same as the area to the left of 2 in Fig. 6-17. The result as shown in Fig. 6-17 is =1-NORMSDIST(2) or 0.02275. Note that the function NORMSDIST has only one parameter and only is used to find cumulative areas under the standard normal curve.

**EXAMPLE 6.9** The time between release from prison and conviction for another crime for individuals under 40 is normally distributed with a mean equal to 30 months and a standard deviation equal to 6 months. The percentage of these individuals convicted for another crime within two years of their release from prison is represented as P(X < 24) times 100. The probability is shown as the shaded area in Fig. 6-18. The event X < 24 is equivalent to The probability P(Z < –1) is shown as the shaded area in Fig. 6-19.

The shaded area shown in Fig. 6-19 is P(Z < –1), and by symmetry P(Z < –1) = P(Z > 1). The probability P(Z > 1) is found using the standard normal distribution table as .5 – .3413 = .1587. That is, P(X < 24) = P(Z < –1) = .1587 or 15.87% commit a crime within two years of their release.

**EXAMPLE 6.10** A study determined that the difference between the price quoted to women and men for used cars is normally distributed with mean $400 and standard deviation $50. To clarify, let X be the amount quoted to a woman minus the amount quoted to a man for a given used car. Then, the population distribution for X is normal with μ = $400 and σ = $50. The percent of the time that quotes for used cars are $275 to $500 more for women than men is given by P(275 < X < 500) times 100. The probability P(275 < X < 500) is shown as the shaded area in Fig. 6-20. The Z value corresponding to X = 275 is and the Z value corresponding to The probability P(–2.5 < Z < 2.0) is shown in Fig. 6-21.

The probability P(–2.5 < Z < 2.0) is found using the standard normal distribution table. The probability is expressed as P(–2.5 < Z < 2.0) = P(–2.5 < Z < 0) + P(0 < Z < 2.0). By symmetry, P(–2.5 < Z < 0) = P(0 < Z < 2.5) = .4938 and P(0 < Z < 2.0) = .4772. Therefore, P(–2.5 < Z < 2.0) = .4938 + .4772 = .971. P(275 < X < 500) = P(–2.5 < Z < 2.0) = .971, or 97.1% of the time the quotes for used cars will be between $275 and $500 more for women than for men.

Practice problems for these concepts can be found at:

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