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 (a) less than 250

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 fewer than 150 would consider working remotely full time? Use the normal approximation if appropriate, and state whether this is an unusual event. So, for this question, let's say that Capital X is the number of employees out of the 350 who would consider working remotely full time. And in this case, it just so happens that X follows a binomial distribution. So, for the binomial distribution, notice that N, that's the sample size, is equal to 350. And P is equal to 0.48 because this, this represents the probability of success, that is the chances that the employee would prefer working remotely full time. So, now this begs the question if the normal distribution is appropriate. So first, what we have to do is check both NP and N multiplied by 1 subtracted by P. So first, NP is equal to 350, multiplied by 0.48, which gives you 168. Now converselyn. Multiplied by 1 subtracted by P. Is 350 multiplied by 1 subtracted by 0.48. And this gives you 182. Now it just so happens that both of these values are greater than 5. And because that's the case, the normal approximation is appropriate. So now we can move on to finding both the mean and standard deviation. Now, the mean, or mu is equal to N multiplied by P. So we already know that that's equal to 168. From there The standard deviation or sigma. Is equal to the square root of. NP multiplied by 1 subtracted by P. So that's equal to the square root of 350. Multiplied by 0.48 and multiplied by 1 subtracted by 0.48. And so this expression simplifies to the square root of 87.36. So now for the continuity correction. So, going back to what the question is asking for, what we're solving for is P of X being less than 150. So for the continuity correction, we subtract 0.5 from 150. So that's X equals 149.5. So, for the Z score that we're about to calculate, what we use in the calculation of the Z score is indeed 149.5. So let me scroll down. More here to open up some space. And now the Z score is equal to 149.5. Subtracted by mu, that's 168, and divided by sigma, that's the square root of. 87.36. So now, evaluating this expression equals approximately 1.975702. And from here, you would find the probability using the normal standard distribution table. Or the standard normal distribution table. When looking at this table, you recognize that our calculated Z score is between two values on the table. Specifically, it's in between -1.97. And -1.98. So, to find the probability that you're looking for, you have to interpolate. So, P of Z being less than. 1.975702. Is equal to. -1.975702. Subtracted by 1.97. Divided by -1.98. Subtracted by -1.97. So now you're going to multiply this term by the difference between. The probabilities of both values. That are Z scores in between. So 4-1.97, that probability is 0. 0244, whereas for negative 1.98, it was 0. 0239 So That's 0.0239, subtracted by 0.0244. And so you're also going to add 0.0244. And so ultimately, your final answer, that is your probability, is equal to approximately 0.0241. So the chances that fewer than 150 employees would consider working remotely full time is 0.0241. So now, is this event unusual? Well, recall that an event is considered to be unusual if its probability is less than 0.05. So, because our calculated probability is indeed less than 0.05, this is an unusual event. And there you have it. So now that completes our final answer. 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

Unusual Events

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