Question

The average employee in the U.S.A. spends μ = 23 minutes commuting to work each day. Assume that the distribution of commute times is normal with a standard deviation of σ = 8 minutes. (a) What proportion of U.S. employees spend less than 15 minutes a day commuting?

(b) What is the probability of randomly selecting an employee who spends more than 35 minutes commuting each day?

Answer 1

Solution :

Given that,

mean = = 23

standard deviation = = 8

P(X<15 ) = P[(X- ) / < (15-23) /8 ]

= P(z <-1 )

Using z table

= 0.1587

proportion=0.1587

(B)P(x >35 ) = 1 - P(x<35 )

= 1 - P[(x -) / < (35-23) / 8]

= 1 - P(z <1.5 )

Using z table

= 1 -  0.9332

probability= 0.0668