Mathematics, 16.02.2021 05:00 jacobballew2019

Determine if the finite correction factor should be used. If so, use it in your calculations when you find the probability. In a sample of 800 gas stations, the mean price for regular gasoline at the pump was $2.816 per gallon and the standard deviation was $0.008 per gallon. A random sample of size 45 is drawn from this population. What is the probability that the mean price per gallon is less than $2.812?

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Answer from: Victoriag2626

0.0003 = 0.03% probability that the mean price per gallon is less than $2.812

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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.

Finite correction factor:

If the sample is more than 5% of the population, instead of Central Limit Theorem, we use the finite correction factor, as the standard deviation of the sampling distribution would be too big.

The mean price for regular gasoline at the pump was $2.816 per gallon and the standard deviation was $0.008 per gallon.

Large sample, so this can be treated as the population data. Then $\mu = 2.816, \sigma = 0.008$

Sample of 45:

45/800 = 0.056 = 5.6% -> Sample more than 5% of the population, so we use the finite correction factor. This means that s will be multiplied by:

$FCF = \sqrt{\frac{N-n}{N-1}}$

In which n is the size of the sample(45) and N is the size of the population (800). So

$FCF = \sqrt{\frac{755}{799}} = 0.9721$

By the Central Limit Theorem, $n = 45, s = 0.9721*\frac{0.008}{\sqrt{45}} = 0.00116$

What is the probability that the mean price per gallon is less than $2.812?

This is the pvalue of Z when X = 2.812. So

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

By the Central Limit Theorem

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

$Z = \frac{2.812 - 2.816}{0.00116}$

Z = -3.45

Z = -3.45 has a pvalue of 0.0003

0.0003 = 0.03% probability that the mean price per gallon is less than $2.812

Answer from: Quest

me! what did you put for this question?

step-by-step explanation:

Answer from: Quest

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