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.

Athletes on Social Issues In a survey of college athletes, 84% said they are willing to speak up and be more active in social issues. You randomly select 25 college athletes. Find the probability that the number who are willing to speak up and be more active in social issues is (a) at least 24

Accepted Answer

All right, hello, everyone. So, this question says, a factory produces light bulbs and 85% of them pass quality control. If a box contains 75 bulbs, what is the probability that at least 70 bulbs plus quality control? Use the normal approximation if appropriate and state whether this is an unusual event. So let's say here that X is the number of bulbs that pass quality control. And X follows a binomial distribution. Where N is equal to 75 and P, that's the probability of success is 0.85. So now to check if the normal approximation is appropriate. To do that, we have to check both NP and N multiplied by 1 subtracted by P. So NP is equal to 75 multiplied by 0.85. Which equals 63.75? On the other hand, N multiplied by 1 subtracted by P is equal to 75 multiplied by 0.15, which gives you 11.25. Because both of these numbers are greater than 5, the normal approximation is appropriate in this case. So now to find the mean and standard deviation. The mean or me. is equal to NP. Which we already know is 63.75. So sigma, that's the standard deviation. is equal to NP multiplied by 1 subtracted by P. That's the square root of 75. Multiplied by 0.85, multiplied by 0.15. And this expression simplifies to the square root of 9.5625. Now to apply the continuity correction. Now, here, what we're looking for is P of X being greater than or equal to 70. So applying the continuity correction for the normal approximation means subtracting 0.5 from 70. So this gives you P of X greater than. 69.5. So now to find the Z score that corresponds to 69.5. So Z is equal to 69.5, subtracted by mu, that's 63.75, and divided by the square root of 9.5625. This ultimately equals 1.8594. So now Using the standard normal table, you would find the corresponding probability for your Z score. But what you'll notice is that for this case, linear interpolation is going to be necessary. Because Z equals 1.8594 falls between. 1.85 and 1.86. So from the standard normal table, P of Z less than 1.85 is equal to 0.9678. Whereas for 1.86, it's 0.9686. So by linear interpolation, P of Z less than 1.8594. Is equal to 1.8594. Subtracted by 1.85. And divided by 1.86 subtracted by 1.85. This is then multiplied by 0.9686, subtracted by 0.9678. And then you add 0.9678. So this probability is ultimately equal to approximately 0.9686. So therefore On to now our final answer or close to it. Because in this case, recall that we're solving for P of Z greater than 1.8594. Which is equal to one subtracted by P of Z less than 1.8594. So this is equal to one subtracted by 0.9686. Which equals 0.0314? And so the probability. That at least 70 bulbs past quality control is 0.0314. And so on to the interpretation. Recall that events are considered unusual if the probability is less than 0.05. Because our calculated value is indeed less than. 0.05. This is an unusual event. Now, in case I misspoke earlier, I just want to clarify that an event is unusual if the probability is less than 0.05, which in this case, it is. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

��app

Key Concepts

Binomial Distribution

Normal Approximation to the Binomial

Probability Calculation

Watch next