The mean lifetime of a tire is 4242 months with a variance of 4949. If 145145 tires are sampled, what is the probability that the mean of the sample would be greater than 42.842.8 months

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Complete Question

The mean lifetime of a tire is 42 months with a variance of 49. If 145 tires are sampled, what is the probability that the mean of the sample would be greater than 42.8 months.

0.083793

Step-by-step explanation:

We would be using the z score formula.

z score formula = z = (x - μ)/σ/√n

where

x is the raw score = 42.8

μ is the population mean = 42

σ is the population standard deviation =

In the above question, we were given variance = 49

Standard deviation = √Variance

= √49

= 7

n = number of samples = 145

z score = z = (x - μ)/σ/√n

= (42.8 - 42)/ (7/√145)

= 0.8/ 0.581318359

= 1.37618

Approximately to 2 decimal places= 1.38

Using the z score table to find the probability.

P(x ≤ 42.8) = P(z = 1.38) = 0.91621

P(x>42.8) = 1 - P(x<42.8)

1 - 0.91621

= 0.083793

Therefore, the probability that the mean of the sample would be greater than 42.8 is 0.083793

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gmt is 4 hrs ahead of edt so just add four hours to 1: 14: 20 (minutes and seconds stay the same)

gmt is 5: 14: 20

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