Bone Density Test A bone mineral density test is used to ident...

Question

Bone Density Test A bone mineral density test is used to identify a bone disease. The result of a bone density test is commonly measured as a z score, and the population of z scores is normally distributed with a mean of 0 and a standard deviation of 1.

c. For a randomly selected subject, find the probability of a bone density test score between -0.67 and 1.29.

c. For a randomly selected subject, find the probability of a bone density test score between -0.67 and 1.29.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says in a bone mineral density test, the test scores are normally distributed with a mean of zero and a standard deviation of 1. Find the probability that a randomly selected subject has a bone density test score between -0.67 and 1.29. We're given 4 possible choices as our answers. For choice A, we have 0.0055. For choice B, we have 0.4789. For choice C, we have 0.6501, and for choice D, we have 0.1709. So, we need to find the probability that a randomly selected subject has a bone density test score between -0.67 and 1.29. So, to do this, we actually want to convert our bone density test score into a Z score. So recall to to do that, we're going to have Z is going to be equal to the quantity of X minus mu in quantity divided by sigma. So this here is just our definition of calculating Z score, where X here is one of our data points, U is the mean, and sigma is the standard deviation. Now, we were given the mean and standard deviation problem, so we'll substitute in those values. So this is going to be equal to the quantity of X minus, and for the mean, we have a value of 0. In quantity, that's gonna be divided by the standard deviation, which we're told the problem is one. Which turns out to be equal to x. So, our bone density test score is already our Z score. And if you call that the Z score is from a normal distribution with a mean of zero and a standard deviation 1, which is exactly what we have here for a bone density test. So, now we need to do is to use our Z tables. And find the probability that Z is going to be less than each of our two values, so the minus 0.67 and 1.29. So, we'll first look at the probability P of Z being less than -0.67, and if we look that up in our Z tables, we see that this is going to be equal to. 0.2514. Next, we'll need to find the cumulative probability, E or Z being less than. 1.29. And again, looking that up in our Z tables, we see that this value is 0.9015. So, to find the probability that a randomly selected subject has a bone density test score between -0.67 and 1.29, we just need to subtract these two cumulative probabilities. That means that the probability E. Of finding Z between. -0.67. And 1.29. Is going to be equal to. The probability that Z is less than 1.29. Minus the probability. That Z is less than minus 0.67. And so substituting in the values, this is going to be equal to 0.9015 minus 0.2514. Which is equal to 0.6501. So this is the answer that we're looking for for this problem. And if we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice C. So I hope this explanation has been useful, and I'll see you in the next video.

Key Concepts

Z Score

Normal Distribution

Probability Calculation

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