Finding the Probability of a Bone Density Test Result Questions

Using the NORM.S.DIST Function to Find the Probability of a Bone Density Test Result

A bone mineral density test can be helpful in identifying the presence or likelihood of osteoporosis, a disease causing bones to become more fragile and more likely to break. The result of a bone density test is commonly measured as a z score. The population of z scores is normally distributed with a mean of 0 and a standard deviation of 1, so these test results meet the requirements of a standard normal distribution; the figure is a graph of these test results. A randomly selected adult undergoes a bone density test. Find the probability that the result is a reading less than 1.27.

We need to find the area in the distribution below z = 1.27.

The area below z = 1.27 is equal to the probability of randomly selecting a person with a bone density test result that is less than 1.27.

Begin with the z score of 1.27 by locating 1.2 in the left column; next find the value in the adjoining row of prob-abilities that is directly below 0.07, as shown in the accompanying excerpt. Table A-2 shows that there is an area of 0.8980 corresponding to z = 1.27. We want the area below 1.27, and Table A-2 gives the cumulative area from the left, so the desired area is 0.8980. Because we have a correspondence between area and probability, we know that the probability of a z score below 1.27 is 0.8980.

Now you will use Excel’s NORM.S.DIST function to find the answer.

1. Open a blank Excel workbook
2. Select cell A1
3. At the top of the screen, click FORMULAS then More Functions, select the Statistical category. Select the NORM.S.DIST function.

1. Complete the dialog box as shown below. Click OK. The NORM.S.DIST function returns a probability of 0.897958.

The probability that a randomly selected person has a bone density test result below 1.27 is 0.8980. Another way to interpret this result is to conclude that 89.80% of people have bone density levels below 1.27.

Using the NORM.S.DIST Function to Find the Probability of a Bone Density Test Result

Using the same bone mineral density test. A randomly selected adult undergoes a bone density test. Find the probability that the result is a reading above -1.00.

We again find the desired probability by finding a corresponding area. We are looking for the area of the region to the right of z = -1.00 that is shaded in the figure.

If we use Table A-2, we should know that it is designed to apply only to cumulative areas from the left. Referring to the page with negative z scores, we find that the cumulative area from the left up to z = -1.00 is 0.1587. Because the total area under the curve is 1, we can find the shaded area by subtracting 0.1587 from 1. The result is 0.8413. Even though Table A-2 is designed only for cumulative areas from the left, we can use it to find cumulative areas from the right.

Using Excel, we need to find the area in the distribution above z = -1.00.

1. Select cell A2
2. Start by typing =1-, do not hit enter.
3. At the top of the screen, click FORMULAS then More Functions, select the Statistical category. Select the NORM.S.DIST function.
4. Complete the dialog box as shown below. Click OK. The NORM.S.DIST function returns a probability of 0.158655. Notice in cell A2 the probability is 0.841345.

Because of the correspondence between probability and area, we conclude that the probability of randomly selecting someone with a bone density reading above -1 is 0.8413 (which is the area to the right of z = -1.00). We might also say that 84.13% of people have bone density levels above -1.00

Using the NORM.S.INV Function to Find the Bone Density Test z Score for a Given Area Under the Normal Curve

Using the same bone density test described in Example 3, we have a standard normal distribution with a mean of 0 and a standard deviation of 1. Find the bone density test score that separates the bottom 2.5% and find the score that separates the top 2.5%.

The required z scores are shown in the figure.

If using Table A-2 to find the z score located to the left, we search the body of the table for an area of 0.025. The result is z = -1.96. To find the z score located to the right, we search the body of Table A-2 for an area of 0.975. (Remember that Table A-2 always gives cumulative areas from the left.) The result is z = 1.96. The values of z = -1.96 and z = 1.96 separate the bottom 2.5% and the top 2.5%.

We will now use Excel to find the answer.

1. We will start with the bottom 2.5%. Select cell A3
2. At the top of the screen, click FORMULAS then More Functions, select the Statistical category. Select the NORM.S.INV function.
3. Complete the dialog box as shown below. Click OK. The NORM.S.INV function returns z = -1.95996.

1. Next, we will find the z score for the top 2.5%. Select cell A4
2. At the top of the screen, click FORMULAS then More Functions, select the Statistical category. Select the NORM.S.INV function.
3. Complete the dialog box as shown below. Click OK. The NORM.S.INV function returns z = 1.959964.

For the population of bone density test scores, 2.5% of the scores are equal to or less than -1.96 and 2.5% of the scores are equal to or greater than 1.96. Another interpretation is that 95% of all bone density test scores are between -1.96 and 1.96.

Using the NORM.DIST Function to Find the Proportion of Women Eligible for Tall Clubs International

The social organization Tall Clubs International has a requirement that women must be at least 70 in. tall. Given that women have normally distributed heights with a mean of 63.8 in. and a standard deviation of 2.6 in., find the percentage of women who satisfy that height requirement.

The figure shows this information:

Women have heights that are normally distributed with a mean of 63.8 in. and a standard deviation of 2.6 in. The shaded region represents the women who satisfy the height requirement by being at least 70 in. tall is 0.0087

We can convert the height of 70 in. to a z score by using as follows:

To use Table A-2, refer to that table with z = 2.38 and find that the cumulative area to the left of z = 2.38 is 0.9913. (Remember, Table A-2 is designed so that all areas are cumulative areas from the left.) Because the total area under the curve is 1, it follows that the shaded area in the figure is 1 – 0.9913 = 0.0087.

We will now use Excel.

1. Select cell A5
2. Start by typing =1-, do not hit enter.
3. At the top of the screen, click FORMULAS then More Functions, select the Statistical category. Select the NORM.DIST function.
4. Complete the dialog box as shown below. Click OK. The NORM.DIST function returns a probability of 0.008548.

Excel’s NORM.DIST function returns a cumulative area of 0.9915. This is the area to the left of the point corresponding to 70 in. You want the area to the right of this point. This area is equal to 1− 0.9915 = 0.0085. The proportion of women taller than 70 in. is 0.0085.

Using the NORM.INV Function to Find the Wechsler IQ Scores That Separate Three Groups

Some educators argue that all students are served better if they are separated into groups according to their abilities. Assume that students are to be separated into a group with IQ scores in the bottom 30%, a second group with IQ scores in the middle 40%, and a third group with IQ scores in the top 30%. The Wechsler Adult Intelligence Scale yields an IQ score obtained through a test, and the scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the Wechsler IQ scores that separate the three groups.

We begin with the graph.

We have entered the mean of 100, and we have identified the x values separating the lowest 30% and the highest 30%. Using Table A-2, we must work with cumulative areas from the left. For the leftmost value of x, the cumulative area from the left is 0.3, so search for an area of 0.3000 in the body of the table to get z = -0.52 (which corresponds to the closest area of 0.3015). For the rightmost value of x, the cumulative area from the left is 0.7, so search for an area of 0.7000 in the body of the table to get z = 0.52 (which corresponds to the closest area of 0.6985). Having found the two z scores, we now proceed to convert them to IQ scores.

We now solve for the two values of x by using

Referring to the figure, we see that the leftmost value of x = 92.2 is reasonable because it is less than the mean of 100. Also, the rightmost value of 107.8 is reasonable because it is above the mean of 100.

We will now use Excel.

1. We will start with the lower boundary of the middle 40%. Select cell A6
2. At the top of the screen, click FORMULAS then More Functions, select the Statistical category. Select the NORM.INV function.
3. Complete the dialog box as shown below. Click OK. The NORM.INV function returns an IQ score equal to 92.13399.

1. Next, we will find the upper boundary of the middle 40%. Select cell A7
2. At the top of the screen, click FORMULAS then More Functions, select the Statistical category. Select the NORM.INV function.
3. Complete the dialog box as shown below. Click OK. The NORM.INV function returns an IQ score equal to 107.866.

The Wechsler IQ scores of 92.2 and 107.8 can be used as cutoff values separating the three groups. Those in the lowest group have IQ scores below 92.2, those in the middle group have IQ scores between 92.2 and 107.8, and those in the highest group have IQ scores above 107.8.

Using the NORM.DIST Function to Find the Probability That a Randomly Selected Male Weighs More Than 156.25

When designing elevators, an obviously important consideration is the weight capacity. An Ohio college student died when he tried to escape from a dormitory elevator that was overloaded with 24 passengers. The elevator was rated for a capacity of 16 passengers with a total weight of 2500 lb. Weights of adults are changing over time and the table shows the mean and standard deviation of males and females.

For the following, we assume a worst-case scenario in which all of the passengers are males (which could easily happen in a dormitory setting). If an elevator is loaded to a capacity of 2500 lb with 16 males, the mean weight of a passenger is 156.25 lb.

a). Find the probability that 1 randomly selected adult male has a weight greater than 156.25 lb.

b). Find the probability that a sample of 16 randomly selected adult males has a mean weight greater than 156.25 lb (so that the total weight exceeds the maximum capacity of 2500 lb).

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a). We seek the area of the green-shaded region in the figure.

We find that the green-shaded area is 0.7432. If using Table A-2, we convert the weight of x = 156.25 lb to the corresponding z score of z = -0.65 as shown here:

We refer to Table A-2 to find that the cumulative area to the left of z = -0.65 is 0.2578, so the green-shaded area in the figure is 1 – 0.2578 = 0.7422. There is a 0.7422 probability that an individual male will weigh more than 156.25 lb.

b). Use the central limit theorem (because we are dealing with the mean of a sample of 16 males, not an individual male).

We can use the normal distribution if the original population is normally distributed. The sample size is not greater than 30, but the original population of weights of males has a normal distribution, so samples of any size will yield means that are normally distributed. Because we are now dealing with a distribution of sample means, we must use the parameters and , which are evaluated as follows:

We want to find the green-shaded area shown the figure.

Using Table A-2, we convert the value of x = 156.25 lb to the corresponding z score of z = -2.61 as shown here:

From Table A-2 we find that the cumulative area to the left of z = -2.61 is 0.0045, so the green-shaded area of the figure is 1 – 0.0045 = 0.9955. The probability that 16 randomly selected males have a mean weight greater than 156.25 lb is 0.9955. There is a 0.9955 probability that 16 randomly selected males will have a mean weight of more than 156.25 lb.

We will now use Excel.

Part a.

1. Select cell A8
2. Start by typing =1-, do not hit enter.
3. At the top of the screen, click FORMULAS then More Functions, select the Statistical category. Select the NORM.DIST function.
4. Complete the dialog box as shown below. Click OK. The NORM.DIST function returns a probability of 0.743182.

There is a 0.7432 probability that an individual male will weigh more than 156.25 lb. The result of 0.7432 from Excel is more accurate than the result found from Table A-2.

Part b.

1. Because we are now dealing with a distribution of sample means, we must use the parameters and , which are evaluated as follows:

1. Select cell A9
2. Start by typing =1-, do not hit enter.
3. At the top of the screen, click FORMULAS then More Functions, select the Statistical category. Select the NORM.DIST function.
4. Complete the dialog box as shown below. Click OK. The NORM.DIST function returns a probability of 0.995509.

There is a 0.9955 probability that 16 randomly selected males will have a mean weight of more than 156.25 lb.

Given that the safe capacity of the elevator is 2500 lb, there is a very good chance (with probability 0.9955) that it will be overweight if is filled with 16 randomly selected males. Given that the elevator was crammed with 24 passengers, it is very likely that the safe weight capacity was exceeded.

Designing Desks for Kindergarten children

You need to obtain new desks for an incoming class of 25 kindergarten students who are all 5 years of age. An important characteristic of the desks is that they must accommodate the sitting heights of those students. (The sitting height is the height of a seated student from the bottom of the feet to the top of the knee.) Table 6-9 lists the parameters for sitting heights of 5-year-old children (based on data from “Nationwide Age References for Sitting Height, Leg Length, and Sitting Height/Height Ratio, and Their Diagnostic Value for Disproportionate Growth Disorders,” by Fredriks et al., Archives of Disease in Childhood, Vol. 90, No. 8).

1. What sitting height will accommodate 95% of the boys? Hint: If the sitting height accommodates 95% of the boys, it accommodates the lowest 95%. Only the boys with sitting heights in the top 5% would not fit. We therefore want the sitting height that separates the lowest 95% from the top 5%, and this corresponds to the sitting height with a cumulative left area of 0.95 under the normal distribution curve.
2. What sitting height is greater than 95% of the means of sitting heights from random samples of 25 boys? Hint: Because we are now working with means from samples of 25 boys we must use the central limit theorem.
3. Based on the preceding results, what single value should be the minimum sitting height accommodated by the desks? Why are the sitting heights of girls not included in the calculations?