Question

According to a particular marketing​ corporation, the per capita consumption of bottled water is 3.8 gallons per month. Assume the standard deviation for this population is 0.85 gallons per month. Consider a random sample of 36 people.

a. What is the probability that the sample mean will be less than 3.6 gallons per​ month?

b. What is the probability that the sample mean will be more than 4.1 gallons per​ month?

c. Identify the symmetrical interval that includes 83​% of the sample means if the true population mean is 3.8 gallons per month.

Answer 1

Part a

We are given µ = 3.8, σ = 0.85, n = 36

We have to find P(Xbar<3.6)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (3.6 – 3.8) / [0.85/sqrt(36)]

Z = -0.2/ 0.141667

Z = -1.41176

P(Z< -1.41176) = P(Xbar<3.6) = 0.07901

(by using z-table)

Required probability = 0.07901

Part b

We are given µ = 3.8, σ = 0.85, n = 36

We have to find P(Xbar>4.1) = 1 – P(Xbar<4.1)

Z = (4.1 - 3.8) / [0.85/sqrt(36)]

Z = 0.3/0.141667

Z = 2.117647

P(Z<2.117647) = P(Xbar<4.1) = 0.982898

(by using z-table)

P(Xbar>4.1) = 1 – P(Xbar<4.1)

P(Xbar>4.1) = 1 – 0.982898

P(Xbar>4.1) = 0.017102

Required probability = 0.017102

Part c

WE have to find sample means for middle 83% area.

Remaining area = 1 – 0.83 = 0.17

Area at left = 0.17/2 = 0.085

Area at right = 0.17/2 = 0.085

Z score for left = -1.3722

Z score for right = 1.3722

(by using z-table)

Xbar at left = µ - Z*(σ/sqrt(n))

We are given µ = 3.8, σ = 0.85, n = 36

Xbar at left = 3.8 - 1.3722*(0.85/sqrt(36))

Xbar at left = 3.60561

Xbar at right = 3.8 + 1.3722*(0.85/sqrt(36))

Xbar at right = 3.9944

Answers: 3.60561 and 3.9944