Question

"One year, consumers spent an average of $22 on a meal at a restaurant. Assume that the amount spent on a restaurant meal is normally distributed with a standard deviation of $4. Complete the following:

a. What is the probability that a randomly selected person spent more than $24 on a restaurant meal? (Express the probability as P(X > $24) and round to four decimal places.)

b. What is the probability that a randomly selected person spent between $13 and $20 on a restaurant meal? (Express the probability as P($13 < X < $20) and round to four decimal places.)

c. Between what two values will the middle 95% of the amounts of cash spent on restaurant meals fall? (Provide the range as X = $a and X = $b, and round to the nearest cent as needed.)"

Answer 1

a) The probability that a randomly selected person spent more than $24 on a restaurant meal is 0.6915. b) The probability that a randomly selected person spent between $13 and $20 on a restaurant meal is 0.2963. c) The middle 95% of the amounts of cash spent on restaurant meals fall between $14.16 and $29.84.

Step-by-step explanation:

a. To find the probability that a randomly selected person spent more than $24 on a restaurant meal, we need to calculate the z-score for $24 and then use the standard normal distribution table to find the corresponding probability. The z-score is found using the formula: z = (x - μ) / σ, where x is the value we are interested in, μ is the mean, and σ is the standard deviation. Plugging in the values, z = (24 - 22) / 4 = 0.5. Looking up the z-score in the standard normal distribution table, we find that the probability corresponding to a z-score of 0.5 is approximately 0.6915. Therefore, the probability that a randomly selected person spent more than $24 on a restaurant meal is 0.6915.

b. To find the probability that a randomly selected person spent between $13 and $20 on a restaurant meal, we need to calculate the z-scores for $13 and $20 and then calculate the area between these two z-scores. The z-score for $13 is z = (13 - 22) / 4 = -2.25, and the z-score for $20 is z = (20 - 22) / 4 = -0.5. Looking up these z-scores in the standard normal distribution table, we find that the corresponding probabilities are 0.0122 and 0.3085, respectively. To find the probability of the range between $13 and $20, we subtract the smaller probability from the larger probability: 0.3085 - 0.0122 = 0.2963. Therefore, the probability that a randomly selected person spent between $13 and $20 on a restaurant meal is 0.2963.

c. To find the range between which the middle 95% of the amounts of cash spent on restaurant meals fall, we need to calculate the z-scores for the lower and upper bounds of the range. The lower bound is the z-score that corresponds to the cumulative probability of (1 - 0.95) / 2 = 0.025. Looking up this z-score in the standard normal distribution table, we find it is approximately -1.96. To find the corresponding cash value, we use the formula: X = μ + (z * σ), where X is the cash value, μ is the mean, z is the z-score, and σ is the standard deviation. Plugging in the values, X = 22 + (-1.96 * 4) = 14.16. The upper bound is found by using the z-score that corresponds to the cumulative probability of (1 + 0.95) / 2 = 0.975, which is approximately 1.96. Using the same formula, the upper bound is X = 22 + (1.96 * 4) = 29.84. Therefore, the middle 95% of the amounts of cash spent on restaurant meals fall between $14.16 and $29.84.