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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 graphis 1.23.

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 area 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,

Areatotheleftofz=PZ<z=0.8907

The z-score with the cumulative probability of 0.8907is obtained from the standard normal table.

From the standard normal table, the area of 0.8907 is observed corresponding to the row value 1.2 and the column value 0.03, which implies that the z-score is 1.23.

Thus, the z-score value of 1.23 has a left-tailed area of 0.8907.

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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

to four decimal places.

Greater than -2.00.

In Exercises 7–10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.

Sampling Distribution of the Sample Variance

a. Find the value of the population variance σ2.

b. Table 6-2 describes the sampling distribution of the sample mean. Construct a similar table representing the sampling distribution of the sample variance s2. Then combine values of s2that are the same, as in Table 6-3 (Hint: See Example 2 on page 258 for Tables 6-2 and 6-3, which describe the sampling distribution of the sample mean.)

c. Find the mean of the sampling distribution of the sample variance.

d. Based on the preceding results, is the sample variance an unbiased estimator of the population variance? Why or why not?

What is the difference between a standard normal distribution and a non-standard normal distribution?

In Exercises 13–20, use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theatre seats, and classroom seats. (Hint: Draw a graph in each case.)

Mean

St.Dev.

Distribution

Males

23.5 in

1.1 in

Normal

Females

22.7 in

1.0 in

Normal

For females, find the first quartile Q1, which is the length separating the bottom 25% from the top 75%.

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

to four decimal places.

Between -3.00 and 3.00.

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