Question

A gasoline tank for a certain car is designed to hold 15.0 gal of gas. Suppose that the variable x = actual capacity of a randomly selected tank has a distribution that is well approximated by a normal curve with mean 15 gal and standard deviation 0.2 gal. (Round all answers to four decimal places.)

(a) What is the probability that a randomly selected tank will hold at most 14.7 gal? P(x ≤ 14.7) =

(b) What is the probability that a randomly selected tank will hold between 14.4 and 15.2 gal? P(14.4 ≤ x ≤ 15.2) = 0.8400

(c) If two such tanks are independently selected, what is the probability that both hold at most 15 gal? P(x ≤ 15) =

Answer 1

Solution :

Given that mean μ = 15 , standard deviation σ = 0.2

(a)
=> P(x <= 14.7) = P((x - μ)/σ <= (14.7 - 15)/0.2)

= P(Z <= -1.5)

= 1 − P(Z < 1.5)

= 1 − 0.9332

= 0.0668

(b)
=> P(14.4 <= x <= 15.2) = P((14.4 - 15)/0.2 <= (x - μ)/σ <= (15.2 - 15)/0.2)

= P(-3 <= Z <= 1)

= P(Z < 1) - P(Z < -3)

= 0.8413 - [1 − P(Z < 3)]

= 0.8413 - [1 − 0.9987]

= 0.8400

(c)
=> P(x <= 15) = P((x - μ)/σ <= (15 - 15)/0.2)

= P(Z <= 0)

= 0.5

=> The probability that both hold at most 15 gallons = (0.5)^2 = 0.25