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In a recent year, the average movie ticket cost $7.89. In a random sample of 50 movie tickets from various areas, what is the probability that the mean cost is between $7.00 and $7.50, given that the population standard deviation is $1.39?

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Final answer:

To calculate the probability that the mean cost of movie tickets in a sample is between $7.00 and $7.50, we determine the standard error, calculate the z-scores for both price limits, and then find the corresponding probabilities using the standard normal distribution. The probability is the difference between these probabilities.

Step-by-step explanation:

To solve this problem, you can use the concepts of the sampling distribution of the sample mean and the Central Limit Theorem (CLT). Given that the population standard deviation (σ) is $1.39 and the sample size (n) is 50, we can assume that the sample mean will be normally distributed based on the CLT as n ≥ 30.

The first step is to calculate the standard error (SE) of the mean, which is the standard deviation (σ) divided by the square root of the sample size (n):
SE = σ / √ n
SE = $1.39 / √ 50
SE ≈ $0.1964

Next, we can find the z-scores for the lower and upper bounds of the movie ticket cost ($7.00 and $7.50 respectively) using the following formula:
z = (X - μ) / SE
Here X is the value for which you want to find the z-score, and μ is the population mean. For $7.00:
z = ($7.00 - $7.89) / $0.1964
z ≈ -4.53
And for $7.50:
z = ($7.50 - $7.89) / $0.1964
z ≈ -1.98

Now, using the standard normal distribution table, we can find the probabilities corresponding to these z-scores and then find the probability that the sample mean falls between $7.00 and $7.50 by finding the difference between the two probabilities.

User Paul Ardeleanu
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