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Find the probability and interpret the results. If convenient, use technology to find the probability. During a certain week the mean price of gasoline was $2.715 per gallon. A random sample of 35 gas stations is drawn from this population. What is the probability that the mean price for the sample was between $2.693 and $2.725 that week? Assume ? = $0.05. The probability that the sample mean was between $2.693 and $2.725 is ___. (Round to four decimal places as needed.) Interpret the results. Choose the correct answer below. A. About 88% of the population of 35 gas stations that week will have a mean price between $2.693 and $2.725. B. About 88% of samples of 35 gas stations that week will have a mean price between $2.693 and $2.725. C. About 12% of the population of 35 gas stations that week will have a mean price between $2.693 and $2.725. Click to select your answer(s).

          Find the probability and interpret the results. If convenient, use technology to find the probability. During a certain week the mean price of gasoline was $2.715 per gallon. A random sample of 35 gas stations is drawn from this population. What is the probability that the mean price for the sample was between $2.693 and $2.725 that week? Assume ? = $0.05.

The probability that the sample mean was between $2.693 and $2.725 is ___.
(Round to four decimal places as needed.)

Interpret the results. Choose the correct answer below.

A. About 88% of the population of 35 gas stations that week will have a mean price between $2.693 and $2.725.

B. About 88% of samples of 35 gas stations that week will have a mean price between $2.693 and $2.725.

C. About 12% of the population of 35 gas stations that week will have a mean price between $2.693 and $2.725.

Click to select your answer(s).

Show more…

Find the probability and interpret the results. If convenient, use technology to find the probability. During a certain week the mean price of gasoline was 2.715 per gallon. A random sample of 35 gas stations is drawn from this population. What is the probability that the mean price for the sample was between2.693 and 2.725 that week? Assume ? =0.05.

The probability that the sample mean was between 2.693 and2.725 is .
(Round to four decimal places as needed.)

Interpret the results. Choose the correct answer below.

A. About 88% of the population of 35 gas stations that week will have a mean price between $2.693 and $2.725.

B. About 88% of samples of 35 gas stations that week will have a mean price between $2.693 and $2.725.

C. About 12% of the population of 35 gas stations that week will have a mean price between $2.693 and $2.725.

Click to select your answer(s).

Added by Shane M.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert David Nguyen

07/29/2022

Step 1

693 and $2.725 using the formula: \[ z = \frac{x - \mu}{\frac{\sigma}{\sqrt{n}}} \] where $x$ is the given value, $\mu = $2.715, $\sigma = 0.05$, and $n = 35$. For $x = $2.693: \[ z_1 = \frac{2.693 - 2.715}{\frac{0.05}{\sqrt{35}}} \] For $x = $2.725: \[ z_2 = Show more…

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Transcript

00:01 Hi, i'm david and i'm here to have janssen your question.

00:03 In the question here, we are going to discuss about the central limit theorem.

00:09 Let me remind you that if the n greater equal to the 30, then the symbol mean x -pile will be approximately to the normal.

00:17 In such a way that if we take the x -bile, we manage the mean divided by standard division overscority of the n, we obtain the standard normal.

00:26 In this question here, we're given the mean equal to the 2.

00:30 Here we have the 2000, 2 .715 and we have given the sigma it will equal to the 0 .05, the n equal to the 35.

00:47 The question asks you find the probability that the sum will be between the 2 .693 and the 2 .725.

01:00 To find this probability, i need to convert the x -par into the z.

01:04 To do it, i need to apply this formula here.

01:08 And then on the left, i will have the 2 .693 and minus the mean, will be the 2 .7 -15, divided by the standard deviation, over square, the end will be the 35.

01:20 Same thing on the right, 2 .7 -25, and then minus the mean will be 2 .715 over the 0 .05, the bandwidth square root of the 35.

01:33 And every compute who have the z will be between 2 .693 minus 2 .715 divided by 0 .05 times squared into the 35 equal to the minus 2 .6 0.

01:50 And on the right 2 .725 minus 2 .715 divided by 0 .05 times square root of the 35 equal to the 1 .18.

02:07 Now to find this probability i need to bring up the z table.

02:11 Let me copy the z table and i will put the table on the right here.

02:20 Let me make it bigger.

02:23 Notice that the table i bring here, it has the negative score with the probability in the left tail...

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