How do I figure this out? (standard deviation, and interpretation)

I roll a die 32 times, and add up the values. The total from those 32 rolls is 139; the average roll is 4.34.

I figure the expected average should be 3.5, the sum of 1 through 6, divided by 6. While 4.34 seems high, I don't know that 32 trials is enough to say anything about it.

I looked up standard deviation, but I don't know enough about the definition of the terms, nor enough about the terms used to define the definitions. I figured it to be 6.8, but I have no idea if that is correct. [(value-expected value)squared, all those values summed and then divided by the number of trials, then the square root of that number.] And then, no idea if that is significant.

Quote: Mosca

I roll a die 32 times, and add up the values. The total from those 32 rolls is 139; the average roll is 4.34.

I figure the expected average should be 3.5, the sum of 1 through 6, divided by 6. While 4.34 seems high, I don't know that 32 trials is enough to say anything about it.

I looked up standard deviation, but I don't know enough about the definition of the terms, nor enough about the terms used to define the definitions. I figured it to be 6.8, but I have no idea if that is correct. [(value-expected value)squared, all those values summed and then divided by the number of trials, then the square root of that number.] And then, no idea if that is significant.

I just saw Corbaccio and asked him about your problem. He told me that your population is a uniform discrete distribution over the range of a to b where a=1 and b=6 inclusive. Then the P(X) = 1/N where N = b - a + 1 or in the instant case, P(X) = 1/6 for all X.

In general the mean of the uniform discrete distribution = E(X) = (a + b) /2 and the variance is Var(X) ={ (b - a + 2) (b - a)} / 12.
Therefore E(X) = 3.5 and Var(X) = 35/12 = 2.91666.

Sample size is n = 32 and sample mean = 4.34

Test the null hypothesis that the population mean is E(X) = 3.5
verses the alternative hypothesis that E(X) not = 3.5. two-tail test. Assume .01 level of significance.

Calculate z = (4.34 - 3.5)/ sq rt (2.91666/32) = .84/ .3019 = 2.78.

Critical values for .01 level are +/- 2.58. Therefore since z = 2.78 > 2.58 Reject the null hypothesis.

Conclusion: accept the alternative hypotheses that E(X) is not = 3.5 .

Ciao,

Matilda