The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 49,360 miles, with a variance of 6,568,969. What is the probability that the sample mean would differ from the population mean by less than 231 miles in a sample of 247 t ires if the manager is correct? Round your answer to four decimal places.

0.9127

Step-by-step explanation:

We solve using this z score formula

z = (x-μ)/√n

where x is the raw score

We are told that:

The sample mean would differ from the population mean by less than 231 miles

μ is the population mean = 49360

Hence,

(x-μ) = 231 Miles

σ is the population standard deviation.

We were given variance = 6,568,969

Standard deviation = √variance

= √6,568,969

= 2563

n = random number of tires = 247 tires

z = 231/256/√247

z =1.41648

Probability value from Z-Table:

P(x < 49591) = 0.92168

≈ 0.9217

The probability that the sample mean would differ from the population mean by less than 231 miles in a sample of 247 tires if the manager is correct is 0.9127

my guess would be b and d, but i didn’t actually get out the pen and paper. lol

step-by-step explanation:

no solution.