At the beginning of the semester, the amount of time a student waits in line at...

At the beginning of the semester, the amount of time a student waits in line at the campus cafeteria is normally distributed with a mean of five minutes and a standard deviation of 1.5 minutes.

If you randomly select 81 students, what is the probability that they have a mean waiting time between 4.7 and 5.8 minutes?

Group of answer choices

.9641

.718

.0359

.282

Solutions

Expert Solution

Solution :

Given that ,

mean = = 5

standard deviation = = 1.5

n = 81

= 5

= / $\sqrt$ n= 1.5 / $\sqrt$ 81=0.1667

P(4.7< <5.8 ) = P[(4.7-5) /0.1667 < ( - ) / < (5.8-5) / 0.1667 )]

= P(-1.80 < Z < 4.79)

= P(Z <4.79 ) - P(Z < -1.80)

Using z table

=1-0.0359

=0.9641

probability=0.9641

orchestra answered 1 year ago

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