The cost of 5 gallons of ice cream has a variance of 64 with a mean of 34 dollars during the summer. What is the probability that the sample mean would differ from the true mean by less than 1.1 dollars if a sample of 38 5-gallon pails is randomly selected

Answer:

0.5587 = 55.87% probability that the sample mean would differ from the true mean by less than 1.1 dollars.

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 normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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

The cost of 5 gallons of ice cream has a variance of 64 with a mean of 34 dollars during the summer.

This means that [tex]\sigma = \sqrt{64} = 8, \mu = 34[/tex]

Sample of 38

This means that [tex]n = 38, s = \frac{8}{\sqrt{38}}[/tex]

What is the probability that the sample mean would differ from the true mean by less than 1.1 dollars ?

P-value of Z when X = 34 + 1.1 = 35.1 subtracted by the p-value of Z when X = 34 - 1.1 = 32.9. So

X = 35.1

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{35.1 - 34}{\frac{8}{\sqrt{38}}}[/tex]

[tex]Z = 0.77[/tex]

[tex]Z = 0.77[/tex] has a p-value of 0.77935

X = 32.9

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{32.9 - 34}{\frac{8}{\sqrt{38}}}[/tex]

[tex]Z = -0.77[/tex]

[tex]Z = -0.77[/tex] has a p-value of 0.22065

0.77935 - 0.22065 = 0.5587

0.5587 = 55.87% probability that the sample mean would differ from the true mean by less than 1.1 dollars.

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