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.

Social Media A survey of Americans found that 55% would be disappointed if Facebook disappeared. You randomly select 500 Americans and ask them whether they would be disappointed if Facebook disappeared. Find the probability that the number who say yes is (c) between 240 and 280, inclusive.

Accepted Answer

All right, hello, everyone. So, this question says, in a recent study, 48% of employees reported that they would consider working remotely full time. If a company randomly selects 350 employees, what is the probability that the number who would consider working remotely full time. Is between 160 and 180 inclusive. Use the normal approximation if appropriate and state whether this is an unusual event. So let's say that capital X is the number of employees out of the 350 who would consider working remotely. Now, X follows a binomial distribution, in which N is equal to 350, and P, the probability of success is 0.48. So first, let's check if the normal approximation applies here. And to do that, we have to check both NP. And N multiplied by 1 subtracted by P. So NP is equal to 350 multiplied by 0.48. This gives you 168. So, on the other hand, N multiplied by 1 subtracted by P is 350. Multiplied by 0.52, since that's one subtracted by 0.48. And this gives you 182. Now, because both of these values are greater than 5, the normal approximation is appropriate here. So now we move on to finding the mean and the standard deviation. Well, recall that the mean, or mu is equal to NP, which we already know is equal to 168. Now the standard deviation sigma. is equal to the square root of NP multiplied by 1 subtracted by P. So that's equal to. 350 multiplied by 0.48. Multiplied by 0.52. And this simplifies to the square root of 87.36. Now we standardize. And we do so by applying the continuity correction. So recall that what we're solving for. Is the probability of X falling between 160? And 180. And X can also be equal to 160 or 180. Applying the continuity correction. Means subtracting 160 by 0.5 and adding 0.5 to 180, and replacing the less than or equal to signs with less than signs. So our new inequality or our new probability rather. Is the probability of X falling between 159.5. And 180.5. So now we convert both of these values to Z scores. Let's say that Z1 is for 159.5. So Z1 It's equal to 159.5, subtracted by mu, that's 168. Divided by The square root of 87.36. This equals approximately 0.9094. So now Z2 is for 180.5, and this is equal to 180.5, subtracted by 168. Divided by the square root of 87.36. And this equals approximately 1.3374. So now We would find the probability. Of Z being less than these respective Z scores, using the standard normal distribution table. But what you'll notice is that these values as they're written, are not found in the standard normal table. Rather, they're between two values that are on the table. Z1, for example, that's negative. 0.9094 is between. -0.90. And 0.91 or negative 0.91. So the probability. Of the less than -0.90 is approximately 0.1841. Whereas for negative 0.91, that's 0.1814. Z2 That's 1.3374 is also between two values. Specifically, it's between. 1.33. And 1.34. So the probability of Z less than 1.33. Is equal to 0.9082. And for 1.34, that's 0.9099. So for both Z values, we are going to have to use linear interpolation. So let's begin with the probability. Of Z less than 0.9094. This would be equal to -0.9094. Subtracted by 0.90. And divided by -0.91, subtracted by 0.90. This term is then multiplied by 0.1814, subtracted by 0.1841. And then you add 0.1841. This ultimately equals approximately 0.1816. And now we do the same thing for the probability of the less than. 1.3374. So by interpolation, that's equal to 1.3374. Subtracted by 1.33. And divided by 1.34 subtracted by 1.33. This term is then multiplied by 0.9099. Subtracted by 0.9082. And then you add 0.9082. Ultimately, this probability equals approximately 0.9095. So now, to find your final answer. So, for the final step, the probability of Z falling between -0.9094 and 1.3374. Is equal to The probability of Z less than 1.3374. Subtracted by The probability of Z less than -0.9094. So this is equal to 0.9095, subtracted by 0.1816, which equals 0.7279. So ultimately, The probability that the number who would consider working remotely full time is between 160 and 180 inclusive is 0.7279. So how would you interpret this value? Well, recall that events are considered unusual if the probability is less than 0.05. But because 0.7279 is greater than 0.05, This is not considered an unusual event. And there you have it. And so with that being said, thank you so very much for watching, and I hope you found this helpful.

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Key Concepts

Binomial Distribution

Normal Approximation to the Binomial

Probability Calculation

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