Question

The time required for an oil change at a certain service station follows a distribution with mean 11.4 minutes and standard deviation 3.2 minutes. A random sample of 40 oil changes is selected.
(a) Explain why the sampling distribution of ¯ x, the mean time of 40 oil changes, is approximately normally.
(b) Find the mean and the standard deviation of the sampling distribution of ¯ x. Round your answers to one decimal place.
(c) What is the probability that the sample mean time of 40 oil changes will exceed 12 minutes? Round your answer to four decimal places.

Answer 1

Solution :

Given that mean μ = 11.4 , standard deviation σ = 3.2 , n = 40

(a) For samples of size 30 or more, the sample mean is approximately normally distributed, with mean μx-bar = μ and standard deviation σx−bar = σ/sqrt(n), where n is the sample size. The larger the sample size, the better the approximation

x-bar ~ N(μx-bar , σx−bar)

(b)
=> mean of the sampling distribution of x-bar is

=> μx-bar = μ = 11.4

=> standard deviation of the sampling distribution of x-bar is

=> σx−bar = σ/sqrt(n) = 3.2/sqrt(40) = 0.5060 = 0.5 (rounded)

(c)
=> P(x-bar > 12) = P((x-bar - μx-bar)/σx−bar > (12 - 11.4)/0.6)

= P(Z > 1)

= 1 - P(Z < 1)

= 1 - 0.8413

= 0.1587