The mean number of miles that people drive between oil changes is 4550 miles with a population standard deviation of 1470 miles. Suppose that 49 drivers are randomly selected. What is the probability that these drivers have a sample mean greater than 4800 miles? Round your answer to four decimal places.

0.1169

Step-by-step explanation:

We solve the question using the z score formula

z = (x-μ)/σ/√n where

x is the raw score = 4800 miles

μ is the population mean = 4550 miles

σ is the population standard deviation = 1470 miles

n is the random number of samples = 49 drivers

Greater than sign = >

For x > 4800 miles

z = 4800 - 4550/1470/√49

= 4800 - 4550/1470/7

= 4800 - 4550/210

= 1.19048

Probability value from Z-Table:

P(x<4800) = 0.88307

P(x>4800) = 1 - P(x<4800)

P(x>4800) = 1 - 0.88307

P(x>4800) = 0.11693

Approximately to 4 decimal places = 0.1169

Therefore, the probability that these drivers have a sample mean greater than 4800 miles is 0.1169

b i think

step-by-step explanation:

not really sure

step-by-step explanation:

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