Standard Normal Distribution. Find the indicated z score. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

Short Answer

Expert verified

The z-score indicated in graph is -1.45.

Step by step solution

01

Given information

The bone density score follows the standard normal distribution with a mean of 0 and a standard deviation of 1.

02

Describe the z-score

The z-score is the distance along the horizontal scale of the standard normal distribution (corresponding to the number of standard deviations above or below the mean). These are the values taken by the standard normal distribution, represented by Z.

03

State the relationship between the area and the probability

As the standard normal curve has an area of 1 enclosed under the curve, the area has a one-to-one correspondence with the probability.

The left-tailed area is equal to the cumulative probabilities, which are obtained by using the standard normal table (Table A-2) for z-scores.

To find the required right-tailed area, subtract the cumulative probabilities from 1.

04

Find the z-score

The area represented is mathematically represented as

Areatotherightofz=PZ>z0.9265=1-PZ<zPZ<z=1-0.9265PZ<z=0.0735

The z-score that has the area 0.0735 to the left of it is obtained from the standard normal table as follows.

From the standard normal table, the area of 0.0735 is observed corresponding to the row value of -1.4 and the column value of 0.05, which implies that the z-score is -1.45.

Thus, the indicated z-score with the right-tailed value of 0.9265 is -1.45.

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Most popular questions from this chapter

Standard Normal DistributionIn Exercises 17–36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers

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