Question

Anticipated consumer demand in a restaurant for free-range steaks next month can be modeled by a normal random variable with mean of 1,000 pounds and a standard deviation of 110 pounds. a. what is the probability that demand will exceed 800 pounds? b. what is the probability that demand will be between 900 and 1,100 pounds? c. the probability is 0.15 that demand will be more than how many pounds?

Answer 1

a) Probability that the demand will exceed 800 pounds = P (x > 800) = 0.9656

b) Probability that the demand will be between 900 and 1100 pounds

= P (900 ≤ x ≤ 1,100) = 0.6372

c) The probability of 0.15 that the demand will be more than 114.4 pounds

Step-by-step explanation

The problem is a normal distribution question.

So, to solve this, we would need to use the standardized score for all these weights.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ

z = standardized score

x = score to be standardized

μ = mean of the distribution = 1,000 pounds

σ = standard deviation of the distribution = 110 pounds

Part A

Probability that the demand will exceed 800 pounds = P (x > 800)

z = standardized score = ?

x = score to be standardized = 800 pounds

μ = mean of the distribution = 1,000 pounds

σ = standard deviation of the distribution = 110 pounds

z = (x - μ)/σ

z = (800 - 1000)/110

z = (-200/110)

z = -1.82

P (x > 800) = P (z > -1.82) = P (z ≤ 1.82) = 0.9656

Note that P (z ≤ 1.82) = 0.9656 was obtained from the normal distribution table or the normal distribution calculator.

Part B

Probability that the demand will be between 900 and 1100 pounds

= P (900 ≤ x ≤ 1,100)

x = 900, 1100

For 900 pounds

z = (x - μ)/σ

z = (900 - 1000)/110

z = (-100/110)

z = -0.91

For 1,100 pounds

z = (x - μ)/σ

z = (1100 - 1000)/110

z = (100/110)

z = 0.91

P (900 ≤ x ≤ 1100) = P (-0.91 ≤ z ≤ 0.91) = P (z ≤ 0.91) - P (z ≤ -0.91)

P (z ≤ 0.91) = 0.8186

P (z ≤ -0.91) = 1 - P (z ≤ 0.91) = 1 - 0.8186 = 0.1814

P (z ≤ 0.91) - P (z ≤ -0.91) = 0.8186 - 0.1814 = 0.6372

Part C

The probability is 0.15 that demand will be more than how many pounds?

P (x > X) = 0.15

P (x > X) = P (z > Z) = 0.15

We need to first find Z from the normal distribution tables

We will need to find the value from the tables that matches 0.15

P (z > Z) = 1 - P (z ≤ Z) = 0.15

P (z ≤ Z) = 1 - 0.15 = 0.85

Z = 1.04 (from the tables)

We can then find the value of X using the formula for standardizing

z = standardized score = 1.04

x = score to be standardized = ?

μ = mean of the distribution = 1,000 pounds

σ = standard deviation of the distribution = 110 pounds

z = (x - μ)/σ

1.04 = (x - 1000)/110

x - 1000 = (1.04) (110)

x - 1000 = 114.4

x = 1000 + 114.4

x = 1114.4 pounds

Hope this Helps!!!