Question

As reported by a recent​ survey, the​ fast-food restaurant with the best​ drive-through time had a mean time spent in the​ drive-through of mu equals 63.7 seconds. Assuming​ drive-through times are normally distributed with sigma equals 2.6 ​seconds, complete parts ​(a) through ​(c) below.

a) what is the probability that a randomly selected car will get through the drive-through in less than 60.2 seconds?

b) what is the probability that a randomly seleced car will sepnd more then 70 seconds in the drive-through?

c) What porportion of cars spend between 60.2 and 70 seconds in the drive-through?

round to four decimal points as needed

Answer 1

Solution :

Given that ,

mean = = 63.7

standard deviation = = 2.6

(a)

P(x < 60.2) = P((x - ) / < (60.2 - 63.7) / 2.6)

= P(z < -1.35)

Using standard normal table,

P(x < 60.2) = 0.0885

Probability = 0.0885

(b)

P(x > 70) = 1 - P(x < 70)

= 1 - P((x - ) / < (70 - 63.7) / 2.6)

= 1 - P(z < 2.42)

= 1 - 0.9922

= 0.0078

P(x > 70) = 0.0078

Probability = 0.0078

(c)

P(60.2 < x < 70) = P((60.2 - 63.7)/ 2.6) < (x - ) / < (70 - 63.7) / 2.6) )

= P(-1.35 < z < 2.42)

= P(z < 2.42) - P(z < -1.35)

= 0.9922 - 0.0885

= 0.9037

Proportion = 0.9037