In Exercises 2–4, the random variable x is normally distribute...

Question

In Exercises 2–4, the random variable x is normally distributed with mean mu= 18 and standard deviation sigma 7.6

Find each probability.

b. P(0 < x < 5)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says consider the random variable Y, which follows a normal distribution with a mean mu equal to 10 and a standard deviation stigma equal to 3.2. Determine the probability that Y falls between 2 and 7. In other words, we need to find the probability P of 2 less than Y, which is less than 7. And we're getting 4 possible choices as our answers. For choice A, we have 0.168. For choice B, we have 0.176. For choice Z, we have 0.674, and for choice D, we have 0.146. So we're asked to find the probability that Y falls between the values of 2 and 7. So, the first thing we want to do is convert those values of 2 and 7 into Z scores. So, we call your formula for finding the Z score, that's going to be Z is equal to quantity of X minus mu in quantity, divided by sigma. Where Z here is our Z score, X is going to be one of our Y values, mu is going to be our mean, and sigma is going to be our standard deviation. Now we're given mu and sigma and the problem, so we actually have all the information we need to calculate our two Z scores. So for the first Z score, we'll call it Z1, and that's going to be equal 2, and here we'll look at the Y value of 2. We'll have the quantity of 2 minus, and from you we're given a value of 10 in the problem, so we'll have 10 N for mu, and we'll take that quantity and then divide that by our standard deviation sigma, which we're given a value of 3.2. And when we evaluate this expression, this is going to be equal to minus 2.5. We'll do something similar for the other value of Y, so we'll have Z2, and that's going to be equal to, and here we'll have for a Y value, we'll have 7, so we'll have 7 minus 10, and that quantity is going to be divided by 3.2. And when we evaluate this expression, this comes out to be. -0.9375. So, the next step is to find the cumulative probabilities corresponding to these two Z scores. And we do that by looking up the values in our standard normal distribution tables. So, when we look up the value for Z1, which is minus 2.5, we see that we have a cumulative probability P of Z less than minus 2.5, is equal to. 0.0062. Now, we tried to look up the next Z-value for Z2, we see that it actually falls in between two of the Z-value entries in our table. So, Z2 falls between the values of Z equal to. -0.94. Which has a cumulative probability. P of Z less than minus 0.94, that is equal to. 0 1736 and the other entry is Z equal to minus 0.93. That corresponds to a cumulative probability P of Z less than minus 0.93, that is equal to. 0 01762. So we're actually going to need to use interpolation to find our cumulative probability or a value for Z2. So that cumulative probability, E of Z. Less than minus 0.9375 is going to be equal to, and here recall your formula for interpolation. So, our first term here is going to be the change in our accumulative probabilities, divided by our change in our Z scores from our table. So we'll have The quantity of 0.1762 minus 0.1736 in quantity, divided by the quantity, here we'll have our change in our Z scores, so we'll have minus 0.93, minus. -0.94 in quantity. And then we need to multiply that fraction. By the difference of RZ scores Z2, which is minus 0.9375. Minus the lower value of our 2. The scores, which is going to be minus 0.94 in quantity. Then we'll have to add to that the cumulative probability or Z equal to minus 0.94, which is going to be 0.1736. And when we evaluate the expression, we see that this is going to be equal to. 0.17. Or 25. So now that we have these two cumulative probabilities, we actually can find the probability that our Y value falls between 2 and 7. So, we're looking for the probability P. Of 2 is less than Y. Which is less than 7. And this is equal to the probability P. Of -2.5. is less than Z, which is less than -0.9375. And that probability is equal to the probability. P of Z less than minus 0.9375. Minus P of Z less than -2.5. And we have calculated those two probabilities, so we'll substitute in those values. So, this is going to be equal to. 0.17425. Minus 0.0062. And when we evaluate this expression, we see that this is gonna be approximately equal to. 0.168. And here we've just round it to three decimal places. So this here is the answer that we are 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 A. So I hope that this explanation has been useful, and I'll see you in the next video.

In Exercises 2–4, the random variable x is normally distributed with mean mu= 18 and standard deviation sigma 7.6

Find each probability.

b. P(0 < x < 5)

Key Concepts

Normal Distribution

Z-scores

Probability Calculation

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