Question

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

Answer 1

Answer:

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

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

$\mu = 100, \sigma = \sqrt{64} = 8, n = 44, s = \frac{8}{\sqrt{44}} = 1.21$

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{98.8 - 100}{1.21}$

$Z = -0.99$

$Z = -0.99$ 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