A racing car consumes a mean of 100 gallons of gas per race with a variance of 64. If 44 racing cars are randomly selected, what is the probability that the sample mean would be greater than 98.8 gallons

83.89% probability that the sample mean would be greater than 98.8 gallons

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error

In this problem, we have that:

The standard deviation is the square root of the variance. So

If 44 racing cars are randomly selected, what is the probability that the sample mean would be greater than 98.8 gallons

This probability is 1 subtracted by the pvalue of Z when X = 98.8. So

By the Central Limit Theorem

has a pvalue of 0.1611

1 - 0.1611 = 0.8389

83.89% probability that the sample mean would be greater than 98.8 gallons

the answer is b

are ya tryin to ask something or is this a statement