Metric Value Average daily food cost $3,200 Standard deviation $450 Distribution type Normal a. Find the probability that a random day has food costs exceeding $3,100. (Round to four decimal places) Answer 1 Question 10 b. Find the probability that daily costs are between $2,800 and $3,500. (Round to four decimal places) Answer 2 Question 10 c. Find the probability that a day has food costs below $2,900. (Round to four decimal places) Answer 3 Question 10 d. If the restaurant wants to identify the 20% of days with lowest food costs for analysis, what should the dollar cutoff be? (Round to two decimal places) Answer 4 Question 10

1. Identify the given values: mean \mu = 3200, standard deviation \sigma = 450, and X = 3100.
2. Calculate the z-score:
z = \frac{X - \mu}{\sigma} = \frac{3100 - 3200}{450} = -\frac{100}{450} \approx -0.2222
3. Use the z-score to find the corresponding probability from the standard normal distribution table:
P(Z > -0.2222) \approx 0.5871
4. The probability that food costs exceed $3100 is approximately 0.5871.

b. [Solution] P(2800 < X < 3500) \approx 0.8290

[Step-by-Step]

1. Identify the given values: mean \mu = 3200, standard deviation \sigma = 450.
2. Calculate the z-scores for X = 2800 and X = 3500:
z_1 = \frac{2800 - 3200}{450} \approx -0.8889
z_2 = \frac{3500 - 3200}{450} \approx 0.6667
3. Use the z-scores to find their corresponding probabilities:
P(Z < -0.8889) \approx 0.1879
P(Z < 0.6667) \approx 0.7470
4. Calculate the probability that daily costs are between $2800 and $3500:
P(2800 < X < 3500) = P(Z < 0.6667) - P(Z < -0.8889) \approx 0.7470 - 0.1879 = 0.8290

c. [Solution] P(X < 2900) \approx 0.2525

[Step-by-Step]

1. Identify the given values: mean \mu = 3200, standard deviation \sigma = 450, and X = 2900.
2. Calculate the z-score:
z = \frac{X - \mu}{\sigma} = \frac{2900 - 3200}{450} = -\frac{300}{450} \approx -0.6667
3. Use the z-score to find the corresponding probability from the standard normal distribution table:
P(Z < -0.6667) \approx 0.2525
4. The probability that food costs are below $2900 is approximately 0.2525.

d. [Solution] C \approx 2876.81

[Step-by-Step]

1. Determine the z-score associated with the lowest 20% of a normal distribution, which is P(Z < z) = 0.20. From standard normal distribution tables, z \approx -0.8416.
2. Use the z-score formula to find the dollar cutoff, C:
C = \mu + z \sigma = 3200 + (-0.8416)(450)
3. Calculate the cutoff:
C = 3200 - 378.19 \approx 2876.81
4. The dollar cutoff for the lowest 20% of days is approximately $2876.81.