Calculating Probability of Sample Mean Differing from True Mean by 2.5 Minutes

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Homework Help Overview

The problem involves calculating the probability that the sample mean of a group of students solving a math problem will differ from the true mean by no more than 2.5 minutes. The context is rooted in statistics, specifically relating to the Central Limit Theorem and sample distributions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the Central Limit Theorem and the formula for z-scores. Questions arise regarding the placement of the 2.5 minutes in the calculations and the necessity of knowing the population mean.

Discussion Status

The discussion is ongoing, with participants offering guidance on how to approach the problem. Some have provided formulas and clarified concepts, while others are seeking further assistance to understand the application of these ideas.

Contextual Notes

There is a mention of a standard deviation of 18 minutes and a sample size of 64 students. The participants are navigating the implications of these values in relation to the problem at hand.

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The time it takes students to solve a certain math problem has a standard deviation of 18 minutes. A random group of 64 students was selected, and they were asked to solve the math problem.

What is the probability that their sample mean will differ from the true mean by no more than 2.5 minutes?

What exactly do I use to determine this problem. Thanks for any help guys.
 
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Central Limit Theorem

[tex]z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}[/tex]

We can give more help if that's not enough.
 
[tex]z=\frac{\bar{x}-\mu}{18/\sqrt{64}}[/tex]

So I don't understand where I would put the 2.5 mins. would go and where I would figure out the population mean.
 
You want [tex]P(|\bar X-\mu|\le 2.5)[/tex]

Divide both sides of [tex]|\bar X-\mu|\le 2.5[/tex] by some number so that you have [tex]P(|Z|\le \text{some constant})[/tex]

Note that you don't need to know the population mean to do this.
 
Any help, anybody? More than happy to figure it out if just given a push in right direction.
 

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