Solved Examples: Measures of Variability



For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for any wrong answer.

For SAT Students
Any question labeled SAT-C is a question that allows a calculator.
Any question labeled SAT-NC is a question that does not allow a calculator.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Show all work.
(1.) Answer these questions, giving reasons.

(a.) In the most recent summer​ Olympics, do you think the standard deviation of the running times for all men who ran the​ 100-meter race would be larger or smaller than the standard deviation of the running times for the​ men's marathon?

(b.) Suppose you have a data set with the weights of all members of a high school soccer team and all members of a high school academic decathlon team​ (a team of students selected because they often answer quiz questions​ correctly).
Which team do you think would have a larger standard deviation of​ weights?


(a.) The standard deviation for the​ 100-meter event would be less.
All the runners come to the finish line within a few seconds of each other.
In the​ marathon, the runners can be quite widely spread after running that long distance.

(b.) The academic decathlon team would have a larger standard deviation of weights.
Academics​ don't require a specific weight to​ succeed, so the distribution of weights should mirror that of the general population.
(2.) JAMB Find the variance of $2, 6, 8, 6, 2$ and $6$

$ A.\;\; \sqrt{5} \\[3ex] B.\;\; \sqrt{6} \\[3ex] C.\;\; 5 \\[3ex] D.\;\; 6 \\[3ex] $

$x$ $f$ $fx$ $x - \bar{x}$ $(x - \bar{x})^2$ $f(x - \bar{x})^2$
$2$ $2$ $4$ $-3$ $9$ $18$
$6$ $3$ $18$ $1$ $1$ $3$
$8$ $1$ $8$ $3$ $9$ $9$
$\Sigma f = 6$ $\Sigma fx = 30$ $\Sigma f(x - \bar{x})^2 = 30$


$ mean = \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] = \dfrac{30}{6} \\[5ex] = 5 \\[3ex] variance = s^2 = \dfrac{\Sigma f(x - \bar{x})^2}{\Sigma f - 1} \\[5ex] = \dfrac{30}{6 - 1} \\[5ex] = \dfrac{30}{5} \\[5ex] = 6 $
(3.)


(4.) JAMB Find the range of $\dfrac{1}{6}, \dfrac{1}{3}, \dfrac{3}{2}, \dfrac{2}{3}, \dfrac{8}{9}$ and $\dfrac{4}{3}$

$ A.\;\; \dfrac{4}{3} \\[5ex] B.\;\; \dfrac{7}{6} \\[5ex] C.\;\; \dfrac{5}{6} \\[5ex] D.\;\; \dfrac{3}{4} \\[5ex] $

$ \dfrac{1}{6}, \dfrac{1}{3}, \dfrac{3}{2}, \dfrac{2}{3}, \dfrac{8}{9}, \dfrac{4}{3} \\[5ex] LCD\;\;of\;\; 6, 3, 2, 9 = 18 \\[3ex] \implies \\[3ex] \dfrac{3}{18}, \dfrac{6}{18}, \dfrac{27}{18}, \dfrac{12}{18}, \dfrac{16}{18}, \dfrac{24}{18} \\[5ex] maximum = \dfrac{27}{18} \\[5ex] minimum = \dfrac{3}{18} \\[5ex] Range = maximum - minimum \\[3ex] = \dfrac{27}{18} - \dfrac{3}{18} \\[5ex] = \dfrac{27 - 3}{18} \\[5ex] = \dfrac{24}{18} \\[5ex] = \dfrac{4}{3} $
(5.)

(6.) JAMB The range of the data $k + 2, k - 3, k + 4, k - 2, k, k - 5, k + 3, k - 1$ and $k + 6$ is

$ A.\;\; 6 \\[3ex] B.\;\; 8 \\[3ex] C.\;\; 10 \\[3ex] D.\;\; 11 \\[3ex] $

$ k + 2, k - 3, k + 4, k - 2, k, k - 5, k + 3, k - 1, k + 6 \\[3ex] Assume\;\; k = 0 \\[3ex] \implies \\[3ex] 0 + 2, 0 - 3, 0 + 4, 0 - 2, 0, 0 - 5, 0 + 3, 0 - 1, 0 + 6 \\[3ex] 2, -3, 4, -2, 0, -5, 3, -1, 6 \\[3ex] minimum = -5 \\[3ex] maximum = 6 \\[3ex] Range = maximum - minimum \\[3ex] = 6 - (-5) \\[3ex] = 6 + 5 \\[3ex] = 11 $
(7.)


(8.) JAMB The range of $4, 3, 11, 9, 6, 15, 19, 23, 27, 24, 21$ and $16$ is

$ A.\;\; 23 \\[3ex] B.\;\; 24 \\[3ex] C.\;\; 21 \\[3ex] D.\;\; 16 \\[3ex] $

$ 4, 3, 11, 9, 6, 15, 19, 23, 27, 24, 21, 16 \\[3ex] minimum = 3 \\[3ex] maximum = 27 \\[3ex] Range = maximum - minimum \\[3ex] = 27 - 3 \\[3ex] = 24 $
(9.)



(10.) JAMB The variance of the scores 1, 2, 3, 4, 5 is

$ A.\;\; 1.2 \\[3ex] B.\;\; 1.4 \\[3ex] C.\;\; 2.0 \\[3ex] D.\;\; 3.0 \\[3ex] $

$x$ $f$ $fx$ $x - \bar{x}$ $(x - \bar{x})^2$ $f(x - \bar{x})^2$
$1$ $1$ $1$ $-2$ $4$ $4$
$2$ $1$ $2$ $-1$ $1$ $1$
$3$ $1$ $3$ $0$ $0$ $0$
$4$ $1$ $4$ $1$ $1$ $1$
$5$ $1$ $5$ $2$ $4$ $4$
$\Sigma f = 5$ $\Sigma fx = 15$ $\Sigma f(x - \bar{x})^2 = 10$


$ mean = \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] = \dfrac{15}{5} \\[5ex] = 3 \\[3ex] variance = s^2 = \dfrac{\Sigma f(x - \bar{x})^2}{\Sigma f - 1} \\[5ex] = \dfrac{10}{5 - 1} \\[5ex] = \dfrac{10}{4} \\[5ex] = 2.5 \\[3ex] $ Because the answer is not in the option, assume that the scores are population scores (rather than sample scores)
In that case, we shall use the population variance

$ population\;\;variance = \sigma^2 = \dfrac{\Sigma f(x - \bar{x})^2}{\Sigma f} \\[5ex] = \dfrac{10}{5} \\[5ex] = 2 \\[3ex] = 2.0 $
(11.) The top seven movies based on DC comic book characters for the U.S. box office as of Fall 2017 are shown in the accompanying​ table, rounded to the nearest million.
Movie Domestic Gross ($ millions)
The Dark Knight (2008) 643
Batman (1989) 547
Superman (1978) 543
The Dark Knight Rises (2012) 487
Wonder Woman (2017) 407
Batman Forever (1995) 366
Superman II (1981) 346

(a.) Sort the domestic gross income from smallest to largest.
(b.) Determine the median and interpret it in context.
(c.) Using the sorted​ data, determine the first quartile and the third quartile.
(d.) Determine the interquartile range and interpret it in context.
(e.) Determine the range of the data.
(f.) Explain why the interquartile range is preferred over the range as a measure of variability.


$ \underline{Domestic\;\;Gross\;\;Income\;(\$\;millions)} \\[3ex] (a.)\;\;sorted\;\;in\;\;ascending\;\;order \\[3ex] 346, 366, 407, 487, 543, 547, 643 \\[5ex] (b.) \\[3ex] n = 7 \\[3ex] Median = 487\;million\;\$ \\[3ex] $ This means that about​ 50% of the top 7 DC movies made more than four hundred and eight seven million dollars.

$ (c.) \\[3ex] n = 7 \\[3ex] 0.25 * 7 = 1.75 \approx 2 \\[3ex] Q_1 = 2nd\;\;value = 366\;million\;\$ \\[3ex] 0.75 * 7 = 5.25 \approx 6 \\[3ex] Q_3 = 6th\;\;value = 547\;million\;\$ \\[5ex] (d.) \\[3ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 547 - 366 = 181\;million\;\$ \\[3ex] $ Interpretation of IQR: The middle​ 50% of the top 7 DC movies had domestic grosses that varied by as much as one hundred and eighty one million dollars.

$ (e.) \\[3ex] max = 643 \;million\;\$ \\[3ex] min = 346 \;million\;\$ \\[3ex] Range = max - min \\[3ex] Range = 643 - 346 = 297 \;million\;\$ \\[3ex] $ (f.) The interquartile range is preferred over the range as a measure of variability because it depends on many observations and is therefore more reliable.
(12.) JAMB If the scores of $3$ students in a test are $5, 6$ and $7$, find the standard deviation of their scores.

$ A.\;\; \dfrac{2}{3} \\[5ex] B.\;\; \dfrac{2}{3} \sqrt{3} \\[5ex] C.\;\; \sqrt{\dfrac{2}{3}} \\[5ex] D.\;\; \sqrt{\dfrac{3}{2}} \\[5ex] $

$x$ $f$ $fx$ $x - \bar{x}$ $(x - \bar{x})^2$ $f(x - \bar{x})^2$
$5$ $1$ $5$ $-1$ $1$ $1$
$6$ $1$ $6$ $0$ $0$ $0$
$7$ $1$ $7$ $1$ $1$ $1$
$\Sigma f = 3$ $\Sigma fx = 18$ $\Sigma f(x - \bar{x})^2 = 2$


$ mean = \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] = \dfrac{18}{3} \\[5ex] = 6 \\[3ex] variance = s^2 = \dfrac{\Sigma f(x - \bar{x})^2}{\Sigma f - 1} \\[5ex] = \dfrac{2}{3 - 1} \\[5ex] = \dfrac{2}{2} \\[5ex] = 1 \\[3ex] standard\;\;deviation = s = \sqrt{s^2} \\[3ex] = \sqrt{1} \\[3ex] = 1 \\[3ex] $ Because the answer is not in the option, assume that the scores are population scores (rather than sample scores)
In that case, we shall use the population standard deviation

$ population\;\;variance = \sigma^2 = \dfrac{\Sigma f(x - \bar{x})^2}{\Sigma f} \\[5ex] = \dfrac{2}{3} \\[5ex] population\;\;standard\;\;deviation = \sigma = \sqrt{\sigma^2} \\[3ex] = \sqrt{\dfrac{2}{3}} ...answer\;\;option \\[5ex] = \dfrac{\sqrt{2}}{\sqrt{3}} \\[5ex] = \dfrac{\sqrt{2}}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\[5ex] = \dfrac{\sqrt{6}}{3} $
(13.)


(14.) JAMB The weights of $10$ pupils in a class are $15\;kg$, $16\;kg$, $17\;kg$, $18\;kg$, $16\;kg$, $17\;kg$, $17\;kg$, $17\;kg$, $18\;kg$ and $16\;kg$
What is the range of this distribution?

$ A.\;\; 1 \\[3ex] B.\;\; 2 \\[3ex] C.\;\; 3 \\[3ex] D.\;\; 4 \\[3ex] $

$ 15, 16, 17, 18, 16, 17, 17, 17, 18, 16 \\[3ex] minimum = 15\;kg \\[3ex] maximum = 18\;kg \\[3ex] Range = maximum - minimum \\[3ex] = 18 - 15 \\[3ex] = 3\;kg $
(15.)


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(18.)