Practice problems for these concepts can be found at:

In many applications, we are concerned with finding the z value or the x value when the area under a normal curve is known. The following examples illustrate the techniques for solving these type problems.

**EXAMPLE 6.11** Find the positive value z such that the area under the standard normal curve between 0 and z is .4951. Figure 6-22 shows the area and the location of z on the horizontal axis. Table 6.3 gives the portion of the standard normal distribution table needed to find the z value.

We search the interior of the table until we find .4951, the area we are given. This area is shown in bold print in Table 6.3. By going to the beginning of the row and top of the column in which .4951 resides, we see that the value for z is 2.58. To find the value of negative z such that the area under the standard normal curve between 0 and z equals .4951, we find the value for positive z as above and then take z to be –2.58.

**EXAMPLE 6.12** The mean number of passengers that fly per day is equal to 1.75 million and the standard deviation is 0.25 million per day. If the number of passengers flying per day is normally distributed, the distribution and ninety-fifth percentile, P_{95}, is as shown in Fig. 6-23. The shaded area is equal to 0.95. We have a known area and need to find P_{95}, a value of X. Since we must always use the standard normal tables to solve problems involving any normal curve, we first draw a standard normal curve corresponding to Fig. 6-23. This is shown in Fig. 6-24.

Using the technique shown in Example 6.11, we find the area .4500 in the interior of the standard normal distribution table and find the value of z to equal 1.645. There is 95% of the area under the standard normal curve to the left of 1.645 and there is 95% of the area under the curve in Fig. 6-23 to the left of P_{95}. Therefore if P_{95} is standardized, the standardized value must equal 1.645. That is, and solving for P_{95} we find P_{95} = 1.75 + .25 × 1.645 = 2.16 million.

The EXCEL solution to Example 6.12 uses the built-in function NORMINV. The expression =NORMINV(0.95,1.75,0.25) entered in any cell gives the answer 2.16121. The first parameter in the function NORMINV is the percentile converted to a decimal, the second is the mean and the third is the standard deviation.

Practice problems for these concepts can be found at:

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