Question

The average amount of money spent for lunch per person in the college cafeteria is $7.13, with a standard deviation of $2.47. Suppose that 8 randomly selected lunch patrons are observed, assuming the distribution of money spent is normal. What is the probability that the average amount spent by these 8 patrons is between $6.50 and $7.80?

A) 0.6687
B) 0.7745
C) 0.3852
D) 0.5324

Answer 1

To find the probability that the average amount spent by 8 randomly selected lunch patrons is between $6.50 and $7.80, we can use the concept of sampling distribution and the Central Limit Theorem. The probability is approximately 0.6687.

Step-by-step explanation:

To find the probability that the average amount spent by 8 randomly selected lunch patrons is between $6.50 and $7.80, we can use the concept of sampling distribution and the Central Limit Theorem.

First, we calculate the standard error of the mean. The standard error is equal to the standard deviation divided by the square root of the sample size. In this case, the standard error is $2.47 / √8 ≈ $0.874.

Next, we convert the given values of $6.50 and $7.80 to z-scores using the formula: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. The z-scores for $6.50 and $7.80 are -0.663 and +1.475, respectively.

Using a standard normal distribution table or a calculator, we can find the probabilities associated with these z-scores. The probability that the average amount spent falls between $6.50 and $7.80 is approximately 0.6687.