The tread life of a particular brand of tire is normally distributed with mean 60,000 miles and standard deviation 3800 miles. Suppose 35 tires are randomly selected for a quality assurance test. Find the probability that the mean tread life from this sample of 35 tires is greater than 59,000 miles. You may use your calculator, but show what you entered to find your answer. Round decimals to the nearest ten-thousandth (four decimal places).

P [ x > 59000} = 0,6057

Step-by-step explanation:

We assume Normal Distribution

P [ x > 59000} = (x - μ₀ ) /σ/√n

P [ x > 59000} =  (59000 - 60000)/ 3800

P [ x > 59000} = - 1000/3800/√35

P [ x > 59000} = - 1000*5,916 /3800

P [ x > 59000} =  - 5916/3800

P [ x > 59000} = - 1,55

We look for p value for that z score n z-table and find

P [ x > 59000} = 0,6057

b 1: 45 am

step-by-step explanation:

its that because if it was in the pm it would be 1345

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step-by-step explanation: