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 (c) between 18 and 22, inclusive.

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 the number of bulbs that pass quality control is between 65 and 69 inclusive. Use the normal approximation if appropriate and state whether this is an unusual event. So first, let's say that X is the number of bulbs that pass the quality control. In this case, X follows a binomial distribution, in which N is equal to 75 and P is equal to 0.85. So first, we need to know if the normal approximation is appropriate. And to do so, we have to check both NP and N multiplied by 1 subtracted by P. So NP is equal to. Oops, 75, that's N. Multiplied by 0.85, and this is equal to 63.75. N multiplied by 1 subtracted by P, on the other hand, is equal to 75, multiplied by 0.15, which equals 11.25. So, because both of these numbers are greater than 5, the normal approximation is appropriate. So now we can find the mean and standard deviation. The mean or mew. If you recall, is equal to NP, which we already know is 63.75. The standard deviation or sigma. Is equal to NP. Multiplied by 1 subtracted by p. Which equals the square root of 75 multiplied by 0.85, multiplied by 0.15. This expression simplifies to the square root of 9.5625. So now we need to apply the continuity correction for the normal approximation. What we're solving for in this case is the probability of X falling between 65. And 69 And X can also be equal to either of those values. So applying the continuity correction means subtracting 65 by 0.5 and adding 0.5 to 69. So the new expression. Is the probability of X falling between 64.5? And 69.5. Keep in mind that the equals of both inequality signs is no longer present after the continuity correction is applied. So now to calculate the Z scores for each value. Starting off with Z1 for X1, which we'll say is equal to 64.5. So Z1 is equal to 64.5, subtracted by mu, that's 63.75. And divided by the square root of 9.5625. This equals approximately 0.2425. And now, let's calculate Z2 for X2, which is equal to 69.5. So 69.5, subtracted by 63.75, and divided by the square root of 9.5625. So this equals approximately 1 point. 859 4 Now to find the probabilities using the standard normal table. Now, when you reference the standard normal table, what you'll notice is that both Z values are actually in between two numbers listed on the table itself. Because of this, we have to do linear interpolation for both probabilities. So let's start with Z1. Now, Z1, which is equal to 0.2425. It is between 0.24. And 0.25 on the standard normal table. Now, P of Z less than 0.24. Is equal to. Approximately 0.5948. Whereas for 0.25, that's 0.5987. Now for Z2 on the other hand. Z2, which is equal to 1.8594. Is between Z values 1.85. And 1.86. So P of Z less than 1.85 is equal to approximately 0.9678. Whereas for 1.86, it's 0.9686. So now for our linear interpolation. Let's start with P of Z less than 0.2425. So this is equal to 0.2425. Subtracted by 0.24. Divided by 0.25, subtracted by 0.24. This expression is then multiplied by. 0.5987. Subtracted by 0.5948. And then you add 0.5948. And this equals approximately 0.5950. So now for our second probability, that's P of Z less than 1.8594. So this is equal to. 1.8594. Subtracted by 1.85. And divided by 1.86, subtracted by 1.85. This term is then multiplied by 0.9686. Subtracted by 0.9678. And then you add 0.9678. And your second probability ultimately is equal to approximately 0.9686. So now to find our answer. So the probability Of Z falling between 0.2425. And 1.8594. is equal to P of Z less than 1.8594. Subtracted by P of. P of Z less than. 0.2425 So this is equal to 0.9686. Subtracted by 0.5950. Which ultimately equals 0.3736? And so the probability That the number of bulbs that pass quality control is between 65 and 69 inclusive is 0.3736. And now for the interpretation. Recall that an event is considered to be unusual if the probability is less than 0.05. Because our calculated probability is greater than 0.05. This is not an unusual event. And there you have it. 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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