The amount of time that a drive-through tank spends on a customer is a random va

Question

The amount of time that a drive-through tank spends on a customer is a random variable with a mean mu = 6.7 minutes and a standard deviation a = 3.0 minutes. If a random sample of 36 customers is observed. find the probability, that mean at the is (a) at most 5.7 minutes (b) more than 7.2 minutes (c) at least 6.7 minutes but less than 7.4 minutes (a) The probability that the mean is at most 5.7 minutes is (Round to four decimal places as needed.) (b) The probability that the mean time is more than 7.2 minutes is (Round to four decimal places as needed.) (c) The probability, that the mean time is between 6.7 minutes and 7.4 minutes is (Round to four decimal places as needed.)

Explanation / Answer

Answer:

a).

standard error = 3.0/sqrt(36) =0.5

z value for 5.7, z =(5.7-6.7)/0.5 =-2

P( x 5.7) = P( z < -2) = 0.0228

b).

z value for 7.2, z =(7.2-6.7)/0.5 =1

P( x >7.2) = P( z >1) = 0.1587

c).

z value for 6.7, z =(6.7-6.7)/0.5 =0

z value for 7.4, z =(7.4-6.7)/0.5 =1.4

P( 6.7<x<7.4) = P( 0<z<1.4)

=P( z <1.4)-P( z< 0)

=0.9192-0.5000

=0.4192

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