I'm having trouble with some stats questions for my psychological testing paper. Have tried emailing the professor but to no avail. Given the standard deviation of 19.8,Identify the range of scores where you would expect to find 68,99.5,& 99.9% of the scores if there were a normal distribution. And then how does the distribution we obtained compare? The mean is 66 for this data set. I've been working on this paper for weeks (it's already over 20 pages long and I think my brain is just wiped out, for the life of me I can't seem to get an answer that makes since. I was sure this wasn't a normal distribution, but according to SPSS it meets the criteria.) If someone could solve it that would be amazing, I know the steps but somehow I'm lost.

data set (Sry for the scribbles, it's like a beautiful mind over here haha

  • 4
    $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. $\endgroup$ – Silverfish Dec 15 '15 at 1:08
  • $\begingroup$ I assume you're in Holden's class? I'm looking for the same answer lol. $\endgroup$ – Ali Dec 15 '15 at 4:32
  • 1
    $\begingroup$ What does the "empirical formula" you were given actually state? $\endgroup$ – Glen_b Dec 15 '15 at 4:50

With the mean being 66 and the standard deviation being 20, the scores on a normal distribution at 68% are 46-86; 95.5% are 26-106, and 99.9% are 6-126. Given this, I looked to see what scores our class got that fell between the scores of 46 and 86 and got 62%. Between 26 and 106, fell 99% of our scores, and 100% of our scores fell between 6 and 126 (as there were no scores below 25, or above 100). No sure if I did this correctly, but these are the answers I got.

| cite | improve this answer | |
  • $\begingroup$ I converted the original to a comment, but since you edited it to respond to the question I now assume this is intended as an answer, so I have undeleted. However, I have removed the part that's in the earlier comment (since that's already under the question). $\endgroup$ – Glen_b Dec 15 '15 at 5:44
  • 2
    $\begingroup$ Leaving aside the somewhat dubious benefit of this exercise, I believe that you're more or less doing the kind of calculation that was wanted. $\endgroup$ – Glen_b Dec 15 '15 at 5:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.