In Exercises 33 and 34, find the indicated probabilities. If c...

Question

In Exercises 33 and 34, find the indicated probabilities. If convenient, use technology to find the probabilities.

The daily surface concentration of carbonyl sulfide on the Indian Ocean is normally distributed, with a mean of 9.1 picomoles per liter and a standard deviation of 3.5 picomoles per liter. Find the probability that on a randomly selected day, the surface concentration of carbonyl sulfide on the Indian Ocean is

a. between 5.1 and 15.7 picomoles per liter.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says, suppose the daily amount of nitrogen dioxide in a city is normally distributed with a mean of 18.6 parts per billion, with the standard deviation of 5.1 parts per billion. What is the probability that on a randomly selected day, the nitrogen dioxide level is between 12.0 and 25.0 parts per billion? And we have four possible choices as our answers. For choice A, we have 0.9022. for choice B, we have 0.0978. For choice C we have 0.8953, and for choice D we have 0.7975. So we're asked to find the probability that on a randomly selected date, the nitrogen dioxide level is between 12.0 and 25.0 parts per billion. So the first thing we want to do is to turn those measurements into a Z score. And so we call that for our Z score, we have Z is equal to the quantity of X minus mu in quantity, divided by sigma. Where Z here is our Z score, X here is going to be our nitrogen dioxide level. You is going to be our mean and sigma is going to be our standard deviation. So we need to calculate 2 Z scores. So the first Z score, which we'll call Z1. is going to be for the 12.0 parts per billion. So we'll have Z1 is equal to, for X, we will have 12.0. Minus 4 mu, that is going to be our mean and we're told in the problem that that is going to be 18.6. And then we'll take that quantity and divide it by our standard deviation, which we're told is 5.1 parts per billion, so we'll divide this by 5.1. And so when we evaluate this expression, this is going to be approximately equal to. -1.2941. We'll have to do the same thing for our other band, we'll have Z2. And this is going to be equal to, and for our nitrogen dioxide level, we'll have 25.0. You'll have minus 18.6, and we're gonna divide that whole quantity by our standard deviation of 5.1. And when we evaluate this expression, this is going to be approximately equal to. 1.25. 49. So now we need to find the probabilities corresponding to Z1 and Z2 using our standard normal distribution table. Now, if we look at Z1, we see that it actually falls between two values in our table. So it falls between Z equal to -1.30. Which has a cumulative probability corresponding to 0.0968. And the other C value is going to be Z is equal to -1.29. And that corresponds. To a probability of 0.0. 985. So, what we're going to do to find our probability corresponding to Z1, we're actually going to have to use um interpolation. So our probability. For Z, less than Z1, which is going to be minus 1.2941. is going to be equal to. And we call that here, we're going to have the change in our probabilities at our table divided by our change in our Z values. So we're gonna have the quantity of 0 point. 0 985. Minus 0.0. 968 in quantity, and that's gonna be divided by the quantity of minus 1.29. 1.30. In quantity. So, we're then going to multiply this by the difference between our Z score that we're looking for, which is Z1, by our lower bound. And so, we'll have the quantity of here in the case Z1, which is gonna be a minus 1.2941. Minus our lower bound, which is going to be -1.30. In quantity. Then we'll have to add to this, the probability for our lower bound. So, the probability for Z equal to -1.30 is going to be 0.0968. And so when we calculate this expression, this is going to be approximately equal to. 0.0. 978. Now we're also going to need to find our probability for Z2. And here again, Z2 falls between two values in our Z table. So it falls between Z equal to. 1.25. And that corresponds to a probability of 0.8944, and the other Z value is going to be Z equal to 1.26, which corresponds to a probability of 0.8962. So, again, we're gonna have to use interpolation to find our cumulative probability for Z2. So, for that probability, we'll have P of Z less than 1.2549. And then it's going to be equal to, and here again, we'll have our change in our probabilities divided by our change in Z values. So we're gonna have In the numerals 0.89. 62 Minus 0.8944. And that quantity is going to be divided by the quantity of 1.26 minus 1.25 in quantity. And then that fraction is going to get multiplied by the quantity of our Z2, which is 1.25 or 9. Minus our lower bound, which is gonna be 1.25. In quantity, and they'll have to add to that the probability for a lower bound, which is 0.8944. And so, evaluating this expression, we're gonna get that our probability here is approximately equal to. 0.8953. So now that we have both of these cumulative probabilities, what we actually want to find is the probability that our nitrogen dioxide levels are between 12.0 and 25.0 parts per billion. So that means we just need to find the probability that our Z score is between our two Z scores that we have here. So, we're really looking for The probability P of minus 1.29 or 1 is less than Z, which is less than 1.25 or 9. And so this is going to be equal to. The probability P of Z less than 1.25. 49 Minus P of Z less than -1.2941. And since we just calculated those two probabilities, we'll substitute in those values, so this is going to be equal to. 0.8953 minus 0.0978. And calculating this difference, this is going to be equal to. 0.7975. So this is the answer that we're looking for for this problem. And if we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice D. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Normal Distribution

Mean and Standard Deviation

Probability Calculation

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