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

The above question is not correctly written

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

1) a. Distributive property

2) subtraction property of equality