Question

The annual per capita consumption of bottled water was 32.6 gallons. Assume that the per capita consumption of bottled water is approximately normally distributed with a mean of 32.6 and a standard deviation of 10 gallons. 

a. What is the probability that someone consumed more than 43 gallons of bottled water? 

b. What is the probability that someone consumed between 30 and 40 gallons of bottled water? 

c. What is the probability that someone consumed less than 30 gallons of bottled water? 

d. 90% of people consumed less than how many gallons of bottled water? 

a. The probability that someone consumed more than 43 gallons of bottled water is _______ (Round to four decimal places as needed.)

Answer 1

a)
Here, μ = 32.6, σ = 10 and x = 43. We need to compute P(X >= 43). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (43 - 32.6)/10 = 1.04

Therefore,
P(X >= 43) = P(z <= (43 - 32.6)/10)
= P(z >= 1.04)
= 1 - 0.8508
= 0.1492

b)
z = (x - μ)/σ
z1 = (30 - 32.6)/10 = -0.26
z2 = (40 - 32.6)/10 = 0.74

Therefore, we get
P(30 <= X <= 40) = P((40 - 32.6)/10) <= z <= (40 - 32.6)/10)
= P(-0.26 <= z <= 0.74) = P(z <= 0.74) - P(z <= -0.26)
= 0.7704 - 0.3974
= 0.3730

c)
P(X <= 30) = P(z <= (30 - 32.6)/10)
= P(z <= -0.26)
= 0.3974

d)
z-value = 1.28

x = 32.6 + 1.28 * 10
x = 45.4

45.4 gallons