Question

.The Quicky Zippy Lube Station has a mean time for an oil change as 12.6 minutes with a standard deviation of 4.15 minutes. (a) What is the probability that an oil change will take more than 16.0 minutes? (b) What is the probability that a random sample of 22 oil changes yields an average between 10.0 and 14.0 minutes?

Answer 1

Solution :

Given ,

mean = = 12.6

standard deviation = = 4.15

P(x >16.0 ) = 1 - P(x< 16.0)

= 1 - P[(x -) / < (16.0 -12.6) / 4.15]

= 1 - P(z < 0.82)

Using z table

= 1 - 0.7939

= 0.2061

probability= 0.2061

(B)

n = 22

= 12.6

=  / n= 4.15 / 22=0.8848

P(10.0<    <14.0 ) = P[(10.0 -12.6) / 0.8848 < ( - ) / < (14.0 -12.6) /0.8848 )]

= P( -2.94< Z <1.58 )

= P(Z <1.58 ) - P(Z <-2.94 )

Using z table

=0.9429-0.0016

=0.9413

probability= 0.9413