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

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

$ 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, $CF$ |
---|---|---|---|---|---|---|

$121.1 - 121.2$ | $6$ | $\dfrac{121.1 + 121.2}{2} = 121.15$ | $121.05 - 121.25$ | $\dfrac{6}{25} = 0.24 = 24\%$ | $6$ | |

$121.3 - 121.4$ | $5$ | $\dfrac{121.3 + 121.4}{2} = 121.35$ | $121.25 - 121.45$ | $\dfrac{5}{25} = \dfrac{1}{5} = 0.2 = 20\%$ | $6 + 5 = 11$ | |

$121.5 - 121.6$ | IIII | $4$ | $\dfrac{121.5 + 121.6}{2} = 121.55$ | $121.45 - 121.65$ | $\dfrac{4}{25} = 0.16 = 16\%$ | $11 + 4 = 15$ |

$121.7 - 121.8$ | $5$ | $\dfrac{121.7 + 121.8}{2} = 121.75$ | $121.65 - 121.85$ | $\dfrac{5}{25} = \dfrac{1}{5} = 0.2 = 20\%$ | $15 + 5 = 20$ | |

$121.9 - 122.0$ | $5$ | $\dfrac{121.9 + 122.0}{2} = 121.95$ | $121.85 - 122.05$ | $\dfrac{5}{25} = \dfrac{1}{5} = 0.2 = 20\%$ | $20 + 5 = 25$ | |

$\Sigma F = 25$ | $\Sigma RF = 1 = 100\%$ |

(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.