PLEASE HELP I CANT GET THESE TYPE OF PROBLEMS A racing car consumes a mean of 101 gallons of gas per race with a variance of 49.

If 43 racing cars are randomly selected, what is the probability that the sample mean would differ from the population mean by greater than 2.4 gallons? Round your answer to four decimal place

The probability that the sample mean would differ from the population mean by greater than 2.4 gallons is 0.86558

Step-by-step explanation:

The given parameters are;

The mean, = 101 gallons

The population mean = μ

The variance, σ² = 49

The number of racing cars, n = 43

The probability that the sample mean would differ from the population mean by greater than 2.4 gallons is given as follows;

P() >2.4 = P(2.4 > >-2.4)

The formula for the test of the mean, Z₀, is given as follows;

Therefore, we have;

Which gives;

P() >2.4 = P(2.25 > Z > -2.25)

(We round up to 2.25 because that is the accuracy of the z-table)

From which we have;

P(Z > -2.25) - P(Z > 2.25) = (1 - P(Z < -2.25)) -  (1 - P(Z < 2.25))

P() >2.4 =  (1 - 0.1222) - (1 - 0.98778) = 0.86558

The probability that the sample mean would differ from the population mean by greater than 2.4 gallons = 0.86558.

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