Approximating Binomial Probabilities In Exercises 19–26, deter...

Question

Approximating Binomial Probabilities In Exercises 19–26, determine whether you can use a normal distribution to approximate the binomial distribution. If you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. If you cannot, explain why and use a binomial distribution to find the indicated probabilities. Identify any unusual events. Explain.

Advancing Research In a survey of U.S. adults, 77% said are willing to share their personal health information to advance medical research. You randomly select 500 U.S. adults. Find the probability that the number who are willing to share their personal health information to advance medical research is (c) between 380 and 390 inclusive.

Accepted Answer

Hello everyone. Let's take a look at this question together. A company survey shows that 81% of employees are satisfied with their current work schedule. If 250 employees are randomly chosen, what is the probability that the number who are satisfied with their work schedule is between 210 and 220 inclusive? Use the normal approximation if appropriate. Is it answer choice? A 0.2351. Answer choice B 0.8705, answer choice C 0.1276, or answer choice D 0.9981. So in order to solve this question, we have to recall how to determine a probability to determine what is the probability that the number of employees out of 250 randomly chosen employees who are satisfied with their work schedule is between 210 and 220. Inclusive if a company survey shows that 81% of employees are satisfied with their current work schedule. And so the first step in solving this problem is to identify the distribution and the parameters, which we will let the variable X be the number of satisfied employees out of 250, and we know X follows a binomial distribution. More specifically, X follows a binomial distribution with the Parameters and equals 250 and P equals 0.81. And now we can check if the normal approximation is appropriate by checking and multiplied by P and N multiplied by 1 minus P, which N multiplied by P is equal to 250 multiplied by 0.81, which equals 20 2.5, and N multiplied by 1 minus P is equal to 250 multiplied by 0.19. which equals 47.5. And looking at these values, we can identify that both are greater than 5. Therefore, normal approximation is appropriate. And now we can find the mean and the standard deviation, which the mean is mu, which is equal to n multiplied by P. So our mean is 202.5, and our standard deviation, which is sigma, is equal to the square root of N. Supplied by P multiplied by 1 minus P, which we can substitute the values into the formula. So we have the square root of 250 multiplied by 0.81, multiplied by 0.19, which is equal to the square root of 38.475. And now we can apply continuity correction, where we know that we want the probability that X is between 210 and 220 inclusive. So then applying continuity correction, we will use the probability that X is between 209.5 and 220.5. And using this information, we can convert to a Z score. Using the formula for the Z score, our first Z score is 209.5 minus 202.5 divided by the square root of 38.475, which gives a value of approximately 1.1285 as the value for our first Z score, and then for our second Z score we have 220.5 minus 202.5 divided by the square root of 38.475, which gives a value of approximately 2.9019 as the value for our second Z score. And now we can find the probability using the standard normal distribution table, starting with our first. score, which is approximately 1.1285, which we know is between 1.12 and 1.13, so the probability of Z being less than 1.12 is approximately 0.8686, and the probability of Z being less than 1.13 is approximately 0.8708. And for our second Z score, which is approximately 2.9019, we know is between 2.90 and 2.91, so the probability of Z being less than 2.90 is approximately 0.9981, and the probability of Z being less than 2.91 is approximately 0.9982. And using this information, we can then use interpolation. Starting with the first Z score, so we have the probability of Z being less than 1.1285, giving us the following formula through interpolation, which when we simplify this, the probability of Z being less than 1.1285 is approximately 0.8705. And then for our second Z score, by interpolation, we have the probability. Of Z is less than 2.9019, which gives us the following formula. So then we can simplify our equation and by interpolation, the probability of Z being less than 2.9019, we have a value of approximately 0.9981. Therefore, the probability of X being between 210 and 220 inclusion. can be rewritten as the probability of Z being between 1.1285 and 2.9019, which is the probability of Z being less than 2.9019, subtracted by the probability of Z being less than 1.1285, which when substituting our values that we calculated by interpolation, we have 0.9981 Contracted by 0.8705, which gives us a value of 0.1276. So the probability that between 210 and 200 employees inclusive are satisfied with their work schedule is approximately 0.1276. And looking at our answer choices, we can identify that answer choice C is the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Binomial Distribution

Normal Approximation to the Binomial

Probability Calculation

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