 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 Dominic Chukwuemeka

# Solved Examples - Introductory Statistics

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.

Attempt all questions.
Show all work.

(1.) The table below shows the voltage measurements from a home in the City of Truth or Consequences, New Mexico for $25$ days.

Day Voltage (volts)
$1$ $121.3$
$2$ $121.1$
$3$ $121.5$
$4$ $121.7$
$5$ $122.0$
$6$ $121.2$
$7$ $121.2$
$8$ $121.3$
$9$ $121.2$
$10$ $121.7$
$11$ $121.6$
$12$ $121.9$
$13$ $121.5$
$14$ $121.4$
$15$ $121.8$
$16$ $121.8$
$17$ $121.2$
$18$ $121.4$
$19$ $121.1$
$20$ $121.9$
$21$ $121.7$
$22$ $121.9$
$23$ $121.9$
$24$ $121.3$
$25$ $121.6$

$(a.)$ Draw a frequency distribution table for the data. Your table should have $5$ classes.
$(b.)$ Compute the statistical properties of the classes.

Maximum voltage value = $122.0$
Minimum voltage value = $121.1$
Range = $122.0 - 121.1 = 0.9$
Number of classes = $5$
Class Width = $\dfrac{0.9}{5} = 0.18 \approx 0.2$

Based on the note, Data Organization; If the class intervals (the $LCI$ and the $UCI$) are decimals rounded to one decimal place, then the difference between the lower class interval of a class and the upper class interval of the previous class is $0.1$

$LCI\:\:of\:\:2nd\:\:Class - UCI\:\:of\:\:1st\:\:Class = 0.1 \\[3ex] 1st\:\:Class = 121.1 - 121.2 \\[3ex] 2nd\:\:Class = 121.3 - 121.4 \\[3ex] UCB\:\:of\:\:1st\:\:Class = \dfrac{UCI\:\:of\:\:1st\:\:Class + LCI\:\:of\:\:2nd\:\:Class}{2} = \dfrac{121.2 + 121.3}{2} = \dfrac{242.5}{2} = 121.25 \\[5ex] LCB\:\:of\:\:1st\:\:Class = \dfrac{LCI\:\:of\:\:1st\:\:Class + UCI\:\:of\:\:previous\:\:Class}{2} \\[5ex]$ We do not have a previous class.
But, assuming we did; the $UCI$ of that class would be $121.2 - 0.2 = 121.0$

$\therefore LCB\:\:of\:\:1st\:\:Class = \dfrac{121.1 + 121.0}{2} = \dfrac{242.1}{2} = 121.05 \\[5ex] Class\:\:Boundary\:\:of\:\:1st\:\:Class = 121.05 - 121.25 \\[3ex]$ The Frequency Distribution Table is constructed as shown:

Voltage Intervals Tally Frequency, $F$ Class Midpoints Class Boundaries Relative Frequency, $RF$ Cumulative Frequency, $CF$
$121.1 - 121.2$ IIII I $6$ $\dfrac{121.1 + 121.2}{2} = 121.15$ $121.05 - 121.25$ $\dfrac{6}{25} = 0.24 = 24\%$ $6$
$121.3 - 121.4$ IIII $5$ $\dfrac{121.3 + 121.4}{2} = 121.35$ $121.25 - 121.45$ $\dfrac{5}{25} = \dfrac{1}{5} = 0.2 = 20\%$ $6 + 5 = 11$
$121.5 - 121.6$ IIII $4$ $\dfrac{121.5 + 121.6}{2} = 121.55$ $121.45 - 121.65$ $\dfrac{4}{25} = 0.16 = 16\%$ $11 + 4 = 15$
$121.7 - 121.8$ IIII $5$ $\dfrac{121.7 + 121.8}{2} = 121.75$ $121.65 - 121.85$ $\dfrac{5}{25} = \dfrac{1}{5} = 0.2 = 20\%$ $15 + 5 = 20$
$121.9 - 122.0$ IIII $5$ $\dfrac{121.9 + 122.0}{2} = 121.95$ $121.85 - 122.05$ $\dfrac{5}{25} = \dfrac{1}{5} = 0.2 = 20\%$ $20 + 5 = 25$
$\Sigma F = 25$ $\Sigma RF = 1 = 100\%$

(2.) The ages of the $2016$ United States presidential candidates from $4$ political parties are:

 $70$ $64$ $45$ $45$ $65$ $63$ $66$ $54$ $62$ $58$ $53$ $61$ $71$ $61$ $45$ $48$ $66$ $53$ $63$ $70$ $55$ $68$ $75$ $65$ $63$

Use $5$ classes to form a grouped data.

(3.) $120$ Nursing majors took a standardized test.
The scores are summarized in the Frequency Table as shown:

Scores Frequency Scores Cumulative Frequency
$160 - 179$ $17$ $Less\:\:than\:\:180$ $17$
$180 - 199$ $20$ $Less\:\:than\:\:200$ $37$
$200 - 219$ $19$ $Less\:\:than\:\:220$ $56$
$220 - 239$ $x$ $Less\:\:than\:\:240$ $70$
$240 - 259$ $17$ $Less\:\:than\:\:260$ $87$
$260 - 279$ $33$ $Less\:\:than\:\:280$ $y$

Calculate the values of $x$ and $y$

$56 + x = 70 \\[3ex] x = 70 - 56 \\[3ex] x = 14 \\[3ex] 87 + 33 = y \\[3ex] 120 = y \\[3ex] y = 120$

(4.) CSEC The cumulative frequency distribution of the volume of petrol needed to fill the tanks of $150$ different vehicles is shown below.

Volume (litres) Cumulative Frequency
$11 - 20$ $24$
$21 - 30$ $59$
$31 - 40$ $101$
$41 - 50$ $129$
$51 - 60$ $150$

(a.) For the class $21 - 30$, determine the
(i) lower class boundary
(ii) class width

(b.) How many vehicles were recorded in the class $31 - 40$?

(c.) A vehicle is chosen at random from the $150$ vehicles. What is the probability that the volume of petrol needed to fill its tank is more than $50.5$ litres? Leave your answer as a fraction.

(d.) Byron estimates the median amount of petrol to be $43.5$ liters. Explain why Byron's estimate is INCORRECT.

(e.) On the partially labelled grid below, construct a histogram to represent the distribution of the volume of petrol needed to fill the tanks of the $150$ vehicles.

(a.) For the class $21 - 30$,

$(i)\:\: LCB = \dfrac{21 + 20}{2} = \dfrac{41}{2} = 20.5 \\[5ex] (ii)\:\: CW = 21 - 11 = 10 \\[3ex]$ Let us re-draw the table to show the frequencies
Volume (litres) Frequency Cumulative Frequency, $CF$
$11 - 20$ $24$ $24$
$21 - 30$ $59 - 24 = 35$ $59$
$31 - 40$ $101 - 59 = 42$ $101$
$41 - 50$ $129 - 101 = 28$ $129$
$51 - 60$ $150 - 129 = 21$ $150$

(b.) The number of vehicles recorded in the $31 - 30$ class is the frequency of that class.
Therefore, $42$ vehicles were recorded in the class $31 - 40$

(c.) Prerequisite to answer (c.) is: Probability
The number of vehicles whose tank is more than $50.5$ litres is the frequency of the $51 - 60$ class
Let $E$ be the event of selecting a vehicle whose tank is more than $50.5$ litres
Let $S$ be the sample space - the total number of vehicles.

$n(E) = 21 \\[3ex] n(S) = 150 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{21}{150} \\[5ex] P(E) = \dfrac{7}{50} \\[5ex]$ (d.) Prerequisite to answer (c.) is: Measures of Center
$\Sigma f = 150 \\[3ex] \dfrac{\Sigma f}{2} = \dfrac{150}{2} = 75 \\[5ex] Begin\:\:from\:\:the\:\:first\:\:class \\[3ex] Keep\:\:adding\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:to\:\:10 \\[3ex] 24 + 35 = 59 \\[3ex] 59 + 42 = 101...stop \\[3ex] Or\:\:better still \\[3ex] CF\:\:of\:\:3rd\:\:class = 101...stop \\[3ex]$ Therefore, the class that contains the median is the $3rd$ class = $31 - 40$
Byron's estimate of $43.5$ is in the $4th$ class = $41 - 50$. That is incorrect.