Mathematics, 23.03.2020 18:30 130005401

A racing car consumes a mean of 114 gallons of gas per race with a standard deviation of 7 gallons. If 46 racing cars are randomly selected, what is the probability that the sample mean would differ from the population mean by greater than 1 gallon

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Answer from: esanchez2002fcb

33.20% probability that the sample mean would differ from the population mean by greater than 1 gallon

Step-by-step explanation:

To solve this question, we need 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 114, \sigma = 7, n = 46, s = \frac{7}{\sqrt{46}} = 1.03$

Either the sample mean differs by 1 gallon or less, or it differs by more than 1 gallon. The sum of the probabilities of these events is decimal 1.

Probability it differs by 1 gallon or less

pvalue of Z when X = 114+1 = 115 subtracted by the pvalue of Z when X = 114-1 = 113.

X = 115

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

By the Central Limit Theorem

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

$Z = \frac{115 - 114}{1.03}$

Z = 0.97

Z = 0.97 has a pvalue of 0.8340

X = 113

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

$Z = \frac{113 - 114}{1.03}$

Z = -0.97

Z = -0.97 has a pvalue of 0.1660

0.8340 - 0.1660 = 0.6680

0.6680 probability it differs from the population mean by 1 gallon or less.

Probability it differs from the population mean by greater than 1 gallon.

0.6680 + p = 1

p = 0.3320

33.20% probability that the sample mean would differ from the population mean by greater than 1 gallon

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