Question

John knows that monthly demand for his product follows a normal distribution with a mean of 2,500 units and a standard deviation of 425 units. Given this, please provide the following answers for John.

a. What is the probability that in a given month demand is less than 3,000 units?

b. What is the probability that in a given month demand is greater than 2,200 units?

c. What is the probability that in a given month demand is between 2,200 and 3,000 units?

d. What is the probability that demand will exceed 5,000 units next month?

e. If John wants to make sure that he meets monthly demand with production output at least 95% of the time. What is the minimum he should produce each month?

Answer 1

Solution :

Given that ,

mean = = 2500

standard deviation = = 425

(a)

P(x < 3000) = P((x - ) / < (3000 - 2500) / 425)

= P(z < 1.1765)

= 0.8803

= 0.8803

Probability = 0.8803

(b)

P(x > 2200) = 1 - P(x < 2200)

= 1 - P((x - ) / < (2200 - 2500) / 425)

= 1 - P(z < -0.7059)

= 1 - 0.2401

= 0.7599

Probability = 0.7599

(c)

P(2200 < x < 3000) = P(x < 3000) - P(x < 2200)

= 0.8803 - 0.2401

= 0.6402

Probability = 0.6402

(d)

P(x > 5000) = 1 - P(x < 5000)

= 1 - P((x - ) / < (5000 - 2500) / 425)

= 1 - P(z < 5.88)

= 1 - 1

= 0

Probability = 0

(e)

P(Z > z) = 95%

1 - P(Z < z) = 0.95

P(Z < z) = 1 - 0.95

P(Z < -1.645) = 0.05

z = -1.645

Using z-score formula,

x = z * +

x = -1.645 * 425 + 2500 = 1801

Minimum = 1801 units