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 (b) more than 360

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 more than 200 are satisfied with their work schedule? Use the normal approximation if appropriate. Is it answer choice A 0.5155, answer choice B, 0.4845, answer choice C 0.6264, or answer choice D 0.3736. So in order to solve this question, we have to recall how to determine a probability so that we can determine the probability that more than 200 employees out of 250 employees that are randomly chosen are satisfied with their work schedule, given that a company survey shows. That 81% of employees are satisfied with their current work schedule. And so the first step in determining this probability is to identify the distribution and the parameters, of 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 N equals 250, and P equals 0.81. And then we can check if the normal approximation is appropriate. By checking N multiplied by P and N multiplied by 1 minus P, where N multiplied by P is equal to 250 multiplied by 0.81, which is 202.5, and N multiplied by 1 minus p is equal to 250 multiplied by 0.19, which is 47.5. And looking at these Values we can identify that both are greater than 5, so then normal approximation is appropriate. And now we can find the mean and the standard deviation where the mean is equal to mu, which is equal to our N multiplied by P. So our mean is 202.5 and our standard deviation is sigma, which is equal to the. root of N multiplied by P, multiplied by 1 minus P, so substituting in the values we have 250 multiplied by 0.81, multiplied by 0.19. So our standard deviation is the square root of 38.475. And now we apply continuity correction for the statement of more than 200. So our probability that we want is the probability of X being greater than 200, and with our continuity correction, our probability statement is the probability of X being greater than 200.5, which we then convert. To a Z score using the formula to calculate the Z score and substituting our values so that Z equals 200.5 minus 200 and 2.5 divided by the square root of 38.475, which gives a value of approximately 0.3224 as our Z score. And now we can find the probability using the standard normal distribution table where we know that our Z score of approximately 0.3224 is between 0.32 and 0.33. So then the probability of Z being less than -0.32 is approximately 0.3745, and the probability of Z being less than -0.33 is approximately. 0.3707 and now we can interpolate using the following formula so that the probability of Z being less than 0.3224 is equal to -0.3224 minus -0.32 divided by 0.33 minus -0.32 multiplied by in parentheses 0.3707 minus 0. 3745 plus 0.3745. And when simplified, we have a value of approximately 0.3736. So then the probability of X being greater than 200 is equal to the probability of Z being greater than 0.3224, which we know is calculated by 1 minus the probability of Z being less than -0.3224. Which is equal to 1 minus 0.3736. So the probability of X being greater than 200 is 0.6264. Therefore, the probability that more than 200 employees are satisfied with their work schedule is approximately 0.6264, which, looking at our answer choices, we can identify is answer choice C. So 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.

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Key Concepts

Binomial Distribution

Normal Approximation to the Binomial

Unusual Events

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