If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples: Measures of Variability

Prerequisite: Introductory Statistics
Calculators:
Vertical Data Entry: Measures of Spread
Horizontal Data Entry: Measures of Spread

Samuel Dominic Chukwuemeka (SamDom For Peace) 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 JAMB and CMAT 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.)


Student $E$ made the highest test score because the score is $2.00$ standard deviations above the mean.
(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.)


(a.) A data score is usual if $-2.00 \le z \le 2.00$
A data value is unusual if the $z-score \lt -2.00$ OR the $z-score \gt 2.00$
The lower $z-score$ boundary is $-2.00$
The upper $z-score$ boundary is $2.00$
This implies that $-2.00$ and $2.00$ are the $z-scores$ that separate the unusual $IQ$ scores from the usual $IQ$ scores

(b.) To find the $IQ$ scores (data values), we need to express them in terms of the $z-scores$.

$ z = \dfrac{x - \mu}{\sigma} \\[5ex] \rightarrow x - \mu = z\sigma \\[3ex] x = z\sigma + \mu \\[3ex] Lower\:\:IQ\:\:score\:\:boundary = -2(19) + 107 = -38 + 107 = 69 \\[3ex] Upper\:\:IQ\:\:score\:\:boundary = 2(19) + 107 = 38 + 107 = 145 \\[3ex] Usual\:\:IQ\:\:scores:\:\: 69 \le IQ \le 145 \\[3ex] Unusual\:\:IQ\:\:scores:\:\: IQ \lt 69 \:\:OR\:\: IQ \gt 145 \\[3ex] $ Therefore $69$ and $145$ are the $IQ$ scores that separate the unusual $IQ$ scores from the usual $IQ$ scores
(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.)


Let us represent this information in a table.
Course Grade Point Credit Hour (CH) Grade Point, (GP) = Point * Credit Hour
Algebra $A$ $4$ $4$ $4 * 4 = 16$
English $B$ $3$ $3$ $3 * 3 = 9$
Biology $B$ $3$ $4$ $3 * 4 = 12$
Chemistry $A$ $4$ $4$ $4 * 4 = 16$
Chinese $C$ $2$ $1$ $2 * 1 = 2$
$\Sigma CH = 16$ $\Sigma GP = 55$


$ GPA = \dfrac{\Sigma GP}{\Sigma CH} \\[5ex] GPA = \dfrac{55}{16} \\[5ex] GPA = 3.4375 \\[3ex] GPA \approx 3.44 $
(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.)



Based on the solution in that link, $m = 3$
(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.)


$ \Sigma x = 5 + 8 + x + 12 + (x + 5) + 10 \\[3ex] \Sigma x = 35 + x + x + 5 \\[3ex] \Sigma x = 40 + 2x \\[3ex] n = 6 \\[3ex] \bar{x} = 10 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \rightarrow \Sigma x = \bar{x} * n \\[3ex] 40 + 2x = 10(6) \\[3ex] 40 + 2x = 60 \\[3ex] 2x = 60 - 40 \\[3ex] 2x = 20 \\[3ex] x = \dfrac{20}{2} \\[5ex] x = 10 \\[3ex] \underline{Check} \\[3ex] x = 10 \\[3ex] x + 5 = 10 + 5 = 15 \\[3ex] Data\:\:values = 5, 8, 10, 12, 15, 10 \\[3ex] \Sigma x = 5 + 8 + 10 + 12 + 15 + 10 = 60 \\[3ex] n = 6 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \bar{x} = \dfrac{60}{6} \\[5ex] \bar{x} = 10 $
(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.)


$ Pass\:\:mark = 5 \\[3ex] Number\:\:of\:\:students\:\:who\:\:made\:\:4 = 5 \\[3ex] Number\:\:of\:\:students\:\:who\:\:made\:\:3 = 1 \\[3ex] Number\:\:of\:\:students\:\:who\:\:made\:\:2 = 3 \\[3ex] Number\:\:of\:\:students\:\:who\:\:failed\:\:the\:\:test = 5 + 1 + 3 = 9 $
(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 $
(25.)


$ 2(4x + 1) = 8 - 3(x - 5) \\[3ex] 8x + 2 = 8 - 3x + 15 \\[3ex] 8x + 3x = 8 + 15 - 2 \\[3ex] 11x = 21 \\[3ex] x = \dfrac{21}{11} \\[5ex] $ Check
$ \underline{LHS} \\[3ex] 2(4x + 1) \\[3ex] x = \dfrac{21}{11} \\[5ex] 2\left(4 * \dfrac{21}{11} + 1\right) \\[5ex] 2\left(\dfrac{84}{11} + 1\right) \\[5ex] 2\left(\dfrac{84}{11}\ + \dfrac{11}{11}\right) \\[5ex] 2\left(\dfrac{84 + 11}{11}\right) \\[5ex] 2\left(\dfrac{95}{11}\right) \\[5ex] \dfrac{190}{11} $ $ \underline{RHS} \\[3ex] 8 - 3(x - 5) \\[3ex] x = \dfrac{21}{11} \\[5ex] 8 - 3\left(\dfrac{21}{11} - 5\right) \\[5ex] 8 - 3\left(\dfrac{21}{11} - \dfrac{55}{11}\right) \\[5ex] 8 - 3\left(\dfrac{21 - 55}{11}\right) \\[5ex] 8 - 3\left(-\dfrac{34}{11}\right) \\[5ex] 8 - -\dfrac{102}{11} \\[5ex] 8 + \dfrac{102}{11} \\[5ex] \dfrac{88}{11} + \dfrac{102}{11} \\[5ex] \dfrac{88 + 102}{11} \\[5ex] \dfrac{190}{11} $
(26.)


$ 4 - m = 2m + 16 \\[3ex] 4 - 16 = 2m + m \\[3ex] -12 = 3m \\[3ex] 3m = -12 \\[3ex] m = -\dfrac{12}{3} \\[5ex] m = -4 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 4 - m \\[3ex] m = -4 \\[3ex] 4 - (-4) \\[3ex] 4 + 4 \\[3ex] 8 $ $ \underline{RHS} \\[3ex] 2m + 16 \\[3ex] m = -4 \\[3ex] 2(-4) + 16 \\[3ex] -8 + 16 \\[3ex] 8 $
(27.)


$ 4(3p + 2) - 7 = 3(p - 2) \\[3ex] 12p + 8 - 7 = 3p - 6 \\[3ex] 12p - 3p = -6 - 8 + 7 \\[3ex] 9p = -7 \\[3ex] p = -\dfrac{7}{9} \\[5ex] $ Check
$ \underline{LHS} \\[3ex] 4(3p + 2) - 7 \\[3ex] p = -\dfrac{7}{9} \\[5ex] 4\left(3 * -\dfrac{7}{9} + 2\right) - 7 \\[5ex] 4\left(-\dfrac{21}{9} + \dfrac{18}{2}\right) - 7 \\[5ex] 4\left(\dfrac{-21 + 18}{9}\right) - 7 \\[5ex] 4\left(-\dfrac{3}{9}\right) - 7 \\[5ex] 4\left(-\dfrac{1}{3}\right) - 7 \\[5ex] -\dfrac{4}{3} - 7 \\[5ex] -\dfrac{4}{3} - \dfrac{21}{3} \\[5ex] \dfrac{-4 - 21}{3} \\[5ex] -\dfrac{25}{3} $ $ \underline{RHS} \\[3ex] 3(p - 2) \\[3ex] p = -\dfrac{7}{9} \\[5ex] 3\left(-\dfrac{7}{9} - 2\right) \\[5ex] 3\left(-\dfrac{7}{9} - \dfrac{18}{9}\right) \\[5ex] 3\left(\dfrac{-7 - 18}{9}\right) \\[5ex] 3\left(-\dfrac{25}{9}\right) \\[5ex] -\dfrac{25}{3} $
(28.)


$ 4 - m = 2m + 16 \\[3ex] 4 - 16 = 2m + m \\[3ex] -12 = 3m \\[3ex] 3m = -12 \\[3ex] m = -\dfrac{12}{3} \\[5ex] m = -4 \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 4 - m \\[3ex] m = -4 \\[3ex] 4 - (-4) \\[3ex] 4 + 4 \\[3ex] 8 $ $ \underline{RHS} \\[3ex] 2m + 16 \\[3ex] m = -4 \\[3ex] 2(-4) + 16 \\[3ex] -8 + 16 \\[3ex] 8 $