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The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 48,637 miles, with a variance of 11,282,880. What is the probability that the sample mean would differ from the population mean by less than 778 miles in a sample of 143 tires if the manager is correct

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MrRoyal

Answer:

[tex]P(x < -778) = 0[/tex]

Step-by-step explanation:

Given

[tex]\bar x = 48673[/tex]

[tex]\sigma^2 = 11282880[/tex]

[tex]n = 143[/tex]

Required

[tex]P(x <- 778)[/tex]

First, we calculate the z score

[tex]z = \frac{x}{\sqrt{\sigma^2}/n}[/tex]

So, we have:

[tex]z = \frac{-778}{\sqrt{11282880}/143}[/tex]

[tex]z = \frac{-778}{3359.0/143}[/tex]

[tex]z = \frac{-778}{23.49}[/tex]

[tex]z = -33.12[/tex]

So:

[tex]P(x < -778) = P(z < -33.12)[/tex]

From z score probability, we have:

[tex]P(x < -778) = 0[/tex]

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