In: Statistics and Probability
studying enough in a four credit hour math class-supposed to be 480 minutes
-list of times in minutes (504, 267, 220, 322, 538, 542, 428, 481, 413, 302, 602)
Find:
-observed value
-predicted z-score for each observed value
-correlation coefficient from linear regression
-is this sample approximately normal? why?
-sample mean
-sample deviation
For the given data, we may enter the observed values and the corresponding Z scores in the form a table:
where, Z score is nothing but the standardized value of observations, by dividing the difference between the observed value and the sample mean by the standard deviation:

Computing the sample mean:


= 419.909
Computing the sample standard deviation:


= 126.396
Hence, the Z score can be obtained using the equation:

Substituting the observed values:
| Observed Value (X) | Z Score (Z) |
| 504 | Z = (504 - 419.909) / 126.396 = 0.67 |
| 267 | Z = (267 - 419.909) / 126.396 = -1.21 |
| 220 | Z = (220 - 419.909) / 126.396 = -1.58 |
| 322 | Z = (322 - 419.909) / 126.396 = -0.77 |
| 538 | Z = (538 - 419.909) / 126.396 = 0.93 |
| 542 | Z = (542 - 419.909) / 126.396 = 0.97 |
| 428 | Z = (428- 419.909) / 126.396 = 0.06 |
| 481 | Z = (481- 419.909) / 126.396 = 0.48 |
| 413 | Z = (413 - 419.909) / 126.396 = -0.05 |
| 302 | Z = (302- 419.909) / 126.396 = -0.93 |
| 602 | Z = (602- 419.909) / 126.396 = 1.44 |