Mathematics, 14.04.2020 23:17 1031kylepoe03

The average price of a gallon of gasoline in Missouri was $2.77 (as of 10-25-10 according to the AAA Daily Fuel Gauge Report). Suppose it is known that σ^2=0.32. What is the probability that a sample of 72 gas stations taken that same week will have a sample mean within $0.15 of the population mean? (Hint: You may need to use as many as 4+ decimal places in your computations to come up with the correct answer!)

A. 0.9756
B. 0.0240
C. 0.0398
D. 0.4176
E. Almost 1 2128

Answers: 2

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Answer from: klivingston1012

A. 0.9756

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(which is the square root of the variance) $\sigma$ , 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:

$\mu = 2.77, \sigma = \sqrt{0.32} = 0.5657, n = 72, s = \frac{0.5657}{\sqrt{72}} = 0.0667$

What is the probability that a sample of 72 gas stations taken that same week will have a sample mean within $0.15 of the population mean?

This is the pvalue of Z when X = 2.77 + 0.15 = 2.92 subtracted by the pvalue of Z when X = 2.77 - 0.15 = 2.62. So

X = 2.92

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

By the Central Limit Theorem

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

$Z = \frac{2.92 - 2.77}{0.0667}$

Z = 2.25

Z = 2.25 has a pvalue of 0.9878

X = 2.62

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

$Z = \frac{2.62 - 2.77}{0.0667}$

Z = -2.25

Z = -2.25 has a pvalue of 0.0122

0.9878 - 0.0122 = 0.9756

Answer from: Quest

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.

Answer from: Quest

unless if u could do my questions, if not i wouldn't want to