group sizes might be quite unequal (a TI-83/84 simulation of this produced 4 1s, 11 2s, 4 3s, and 7
4s).Solutions to Cumulative Review Problems
The dataset has an outlier at 85. Because the mean is not resistant to extreme values, it tends to be
pulled in the direction of an outlier. Hence, we would expect the mean to be larger than the median.
- Parameters are values that describe populations, and statistics are values that describe samples.
Hence, the mean and standard deviation of the pulse rates of Pamela’s sample are statistics , and the
predicted mean and standard deviation for the entire school are parameters . - Putting the numbers in the calculator and doing 1-Var Stats , we find that x – = 29.64, s = 11.78,
Q1 = 23, Med = 27, and Q3 = 35.
(a) The interquartile range (IQR) = 35 – 23 = 12, 1.5(IQR) = 1.5(12) = 18. So the boundaries
beyond which we find outliers are Q1 – 1.5(IQR) = 23 – 18 = 5 and Q3 + 1.5(IQR) = 35 + 18 = - Because 55 is beyond the boundary value of 53, it is an outlier, and it is the only outlier.
(b) The usual rule for outliers based on the mean is ± 3s . ± 3s = 29.64 ± 3(11.78) = (-5.7,
64.98). Using this rule there are no outliers since there are no values less than –5.7 or greater
than 64.98. Sometimes ± 2s is used to determine outliers. In this case, ± 2s = 29.64 ± 2
(11.78) = (6.08, 53.2) Using this rule, 55 would be an outlier. - For the given data, = 29.64 and s = 11.78. Hence,
Note that in doing problem #3, we could have computed this z -score and observed that because it is
larger than 2, it represents an outlier by the x – ± 2s rule that is sometimes used.
The problem is referring to the coefficient of determination —the proportion of variation in one
variable that can be explained by the regression of that variable on another.