A racing car consumes a mean of 99 gallons of gas per race with a standard deviation of 3 gallons. If 37 racing cars are randomly selected, what is the probability that the sample mean would be greater than 100.2 gallons? Round your answer to four decimal places.

The probability that the sample mean consumption of gas is greater than 100.2 gallons given a random sample of 37 racing cars is approximately 0.0074 or 0.74 percent.

Explanation:

This question is about calculating probabilities using sample means and based on normal distribution. This case pertains to the field of statistics. Here, the problem can be addressed through Z scores which allow understanding of how far a data point is from the mean in terms of standard deviations.

Step 1: We need to calculate the standard error which equals standard deviation divided by the square root of the sample size (SE = sd/√n). Our given values are sd = 3 gallons and n = 37. So, SE = 3/√37= 0.492.

Step 2: We will calculate the Z score. Z = (X- µ) / SE, where X is the sample mean and µ is the population mean. Plugging in , we get Z = (100.2 - 99) / 0.492 = 2.44.

Step 3: We have to find the probability that Z is greater than 2.44. Using standard Z-tables or a calculator providing probabilities for the normal distribution, the probability of observing a Z score less than 2.44 is about 0.9926. Therefore, the probability that Z is greater than 2.44 equals 1-0.9926= 0.0074.

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