Mathematics: 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).

It’s the first 1 this to easy I’m in high school...

The tread life of a particular brand of tire is

kenz2797

30.01.2022

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

0,0(0 оценок)