helped me gett his answer and it was wrong. Please help Suppose Abby tracks her

Question

helped me gett his answer and it was wrong. Please help

Suppose Abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. Assume the travel times are normally distributed with a standard deviation of 10.3 min. Determine the travel time x such that 17.36% of the 60 days have a travel time that is at leastx You may find this standard normal distribution table or this list of software manuals useful. Give your answer rounded to two decimal places min x = 144.89

Explanation / Answer

Answer to the question is as follows:

n = 60

Mean = 35.6

Stdev = 10.3

Now, P(X>x) = .1736

P(Z> (x-35.6)/(10.3/sqrt(60)) = .1736

(x-35.6)/(10.3/sqrt(60)) = .94

x = 35.6+ .94*10.3/sqrt(60)

x=36.85

So, x = 36.85 minutes is travel time sucg that 17.36% of the 60days have a travel time of alteast x

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