AP Statistics 2017

(Marvins-Underground-K-12) #1
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




  1.          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.



  2. 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 .

  3. 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 =

  4. 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.

  5. 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.



  1.          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.



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