Practice problems for these concepts can be found at:

Figure 6-25 shows the binomial distribution for X, the number of heads to occur in 10 tosses of a coin. This distribution is shown as the shaded area under the histogram-shaped figure. Superimposed upon this binomial distribution is a normal curve. The mean of the binomial distribution is μ = np = 10 × .5 = 5 and the standard deviation of the binomial distribution is The normal curve, which is shown, has the same mean and standard deviation as the binomial distribution. When a normal curve is fit to a binomial distribution in this manner, this is called the *normal approximation to the binomial distribution*. The shaded area under the binomial distribution is equal to one and so is the total area under the normal curve.

The normal approximation to the binomial distribution is appropriate whenever np ≥ 5 and nq ≥ 5.

**EXAMPLE 6.13** Using the table of binomial probabilities, the probability of 4 to 6 heads inclusive is as follows: P(4 ≤ X ≤ 6) = .2051 + .2461 + .2051 = .6563. This is the shaded area shown in Fig. 6-26.

The normal approximation to this area is shown in Fig. 6-27. To account for the area of all three rectangles, note that X must go from 3.5 to 6.5. The area under the normal curve for X between 3.5 and 6.5 is found by determining the area under the standard normal curve for Z between This area is 2 × .3289 = .6578. Note that the approximation, .6578, is very close to the exact answer, .6563.

**EXAMPLE 6.14** To aid in the early detection of breast cancer, women are urged to perform a self-exam monthly. Thirty-eight percent of American women perform this exam monthly. In a sample of 315 women, the probability that 100 or fewer perform the exam monthly is P(X ≤ 100). Even Minitab has difficulty performing this extremely difficult computation involving binomial probabilities. However, this probability is quite easy to approximate using the normal approximation. The mean value for X is μ = np = 315 × .38 = 119.7 and the standard deviation is A normal curve is constructed with mean 119.7 and standard deviation 8.61. In order to cover all the area associated with the rectangle at X = 100 and all those less than 100, the normal curve area associated with x less than 100.5 is found. Figure 6-28 shows the normal curve area we need to find.

The area under the normal curve in Fig. 6-28 is the same as the area under the standard normal curve to the left of Z = –2.23. The area to the left of Z = –2.23 is given by the STATISTIX output shown in Fig. 6-29. The pull-down **Statistics ⇒ Probability functions** gives the dialog box shown in Fig. 6-29. The function chosen is Z 1-Tail(x), –2.23 is entered for X, and Go gives the results highlighted in the figure.

Practice problems for these concepts can be found at:

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