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Numerade Educator

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Problem 13 Easy Difficulty

In the second quarter of $2015,$ the average movie ticket cost $\$ 8.61 .$ In a random sample of 50 movie tickets from various areas, what is the probability that the mean cost exceeds $\$ 8.00,$ given that the population standard deviation is $\$ 1.39 ?$

Answer

0.9990

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Video Transcript

According to this problem, the average movie ticket in the second quarter of 2015 was $8.61. So here we're talking about the population of movie tickets. We also know from the given information that the population standard deviation is a dollar 39. And in this problem, we're going to do a random sample of 50 movie tickets. So from that population, we're drawing a sample in the sample size is going to be 50 tickets. And the question is asking us, What is the probability that the mean movie ticket of these 50 that we have selected is going to be exceeding, which is the same thing is greater than $8. So because we're talking about a sample and the fact that we're dealing with a probability involving the average being greater than $8 we're applying the central limit there and with Central Limit Theorem, we know that the average of the means is going to be the same as the average of the population, and in this case it's $8.61. We also know the standard error of the mean where the standard deviation of the means is equal to the standard deviation of the population divided by the square root of end. And in this case, that would be a dollar 39 divided by the square root of 50. Now we're going to have to draw a bell shaped curve to represent the situation going on. So our bell shaped curve is going to have the average of the center, and our average is $8 in 61 cents. And we're trying to determine the probability that the average is greater than $88 would be somewhere to the left of that and where you're talking going greater than so the next thing we would have to do is we would have to calculate the Z score associated with $8 so we're going to do. The Z score is $8 minus the average of $8.61 divided by the standard deviation, which in this case is 1.39 divided by the square root of 50. And if you get your calculator out, that's approximately and negative 3.10 so we could say on our bell curve that $8 in the negative, 3.10 are equivalent. So that means we can go back. And we can rewrite the original problem in terms of Z scores and say that would be the same as the probability that the Z score is greater than negative 3.1 sirrah. And that's the same thing is saying one minus the probability that Z is less than negative 3.10 You would then go to the standard normal table in the back of your textbook, and you would find the probability of Z being less than negative 0.3 or 3.10 is 0.10 And when we do distraction, we get an answer of 0.99 nine zero. So, to recap, the question is actually stating, What's the probability that the average a movie ticket when we take a sample of 50 turns out to be a greater than $8 and the answer would be 0.9990