Question

Consider a distribution of student scores that is Normal with a mean of 288 and a standard deviation of 38.

1. What is the normalized value (z-score) of a score of 300?
2. What is the proportion of students with scores greater than 300?
3. What is the proportion of students with scores between 290 and 320?
4. Using the 68-95-99.7 rule, what are the two scores symmetrically placed around the mean that would include 68% of the observations?

Answer 1

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The normal distribution has params of Mean = 288 and stdev of 38

1. Normalized score of 300 is (300-288)/38 = .316

2. Proportion of people who had scores greater than 300 is : P(X>300) = P(Z>.316) = .6239

3. Proportion of scores between 290 and 320 is P(Z< .842) - P(Z<..0526) =

=.800-.521

= .279

4. So, by the  68-95-99.7 rule you would have to go 1 deviation away from mean to cover 68% of the data, symmetrically

So, 1 deviation away from mean is 288 +/- 38 or 260 and 326

So, between the symmetrical scores of 260 and 326 would we have 68% of the observations.