Question

One year customers spent an average of $22 on a meal at a resturant. Assume that the amount spent on a resturant meal is normally disttibuted and that the standard deviation is $3 ,

What is the probability that a randomly selected person spent between $14 and $20?(round to four decimal places as needed)

Answer 1

To find the probability that a randomly selected person spent between $14 and $20 on a meal, we need to calculate the z-scores for these values and use the standard normal distribution table. The probability is approximately 0.2475.

Step-by-step explanation:

To find the probability that a randomly selected person spent between $14 and $20 on a meal, we need to calculate the z-scores for these values and then use the standard normal distribution table.

First, we need to calculate the z-scores:

Z-score for $14:

Z = (X - μ) / σ

Z = (14 - 22) / 3 = -2.67

Z-score for $20:

Z = (X - μ) / σ

Z = (20 - 22) / 3 = -0.67

Next, we look up the corresponding cumulative probabilities for these z-scores in the standard normal distribution table.

Cumulative probability for z = -2.67 is 0.0039 (rounded to four decimal places).

Cumulative probability for z = -0.67 is 0.2514 (rounded to four decimal places).

To find the probability between $14 and $20, we subtract the cumulative probability for $14 from the cumulative probability for $20:

P($14 < X < $20) = 0.2514 - 0.0039 = 0.2475 (rounded to four decimal places).