Calculating Probability for Test Scores: Mean 2.725, SD 1.329, 61 Students

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Statistics Need Help !1

I am doing review for a test and i still can not get this problem..HELP Please

The scores on a test had a mean of 2.725 and a standard deviation of 1.329. A teacher had 61 students take the test. Although the students were not random sample, the teacher considered the students to be typical of all the national students. what us the probability that the students achieved an average score of at least 3?

Please help!
 
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I'm going to assume that the distribution is a normal distribution.

Have you studied how to find probabilities in a normal distribution? Have you learned about finding probabilities in a standardized normal distribution (ie mean = 0, std dev = 1)? Do you know how to convert your distribution to a standardized normal distribution?
 


The point of monacom08's problem is to use the CLT to calculate

<br /> P(\overline X \ge 3) = 1 - P(\overline X &lt; 3)<br />

so the first step is to find the sampling distribution of \overline X.
 


Try setting up a solution using Chebycheff's Inequality:

P(|Y - µ| >= kσ) <= 1/k^2

We know that µ is the mean, which equals 2.725, and that σ is the square root of variance (i.e., the standard deviation, which is 1.329).
 


Look at this part of the problem: "what us the probability that the students achieved an average score of at least 3? "

The question asks about the chance the average score is a certain size, not the percentage of individual scores that are a certain size.
 


statdad said:
The question asks about the chance the average score is a certain size, not the percentage of individual scores that are a certain size.

Is this in regards to what I wrote? I am very much a neophyte of statistics, especially compared to you, but how does the number of students and the idea of a percentage achieving an average located above a certain number change the problem? Is Chebysheff's Inequality a viable means to a solution?
 
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