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 Central Tendency

Prerequisite: Introductory Statistics
Calculators: Measures of Center

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.) ACT To determine a student's overall test score for the semester, Ms. Ackerman deletes the lowest test score and calculates the average of the remaining test scores.
Niels took all $5$ tests and earned the following test scores in Ms. Ackerman's class this semester: $62, 78, 83, 86, \:\:and\:\: 93$.
What overall test score did Niels earn in Ms. Ackerman's class this semester?

$ A.\:\: 77.5 \\[3ex] B.\:\: 80.4 \\[3ex] C.\:\: 83.4 \\[3ex] D.\:\: 85.0 \\[3ex] E.\:\: 85.5 \\[3ex] $

$ Lowest\:\:score = 62 \\[3ex] Delete\:\: 62 \\[3ex] Average\:\:of\:\:remaining\:\:scores = \dfrac{78 + 83 + 86 + 93}{4} = \dfrac{340}{4} = 85 \\[5ex] \therefore Overall\:\:test\:\:score = 85.0 $
(2.) JAMB Find the mean of the following $24.57, 25.63, 25.32, 26.01, 25.77$

$ A.\:\: 25.12 \\[3ex] B.\:\: 25.30 \\[3ex] C.\:\: 25.46 \\[3ex] D.\:\: 25.50 \\[3ex] E.\:\: 25.73 \\[3ex] $

$ x = 24.57, 25.63, 25.32, 26.01, 25.77 \\[3ex] \Sigma x = 24.57 + 25.63 + 25.32 + 26.01 + 25.77 \\[3ex] \Sigma x = 127.3 \\[3ex] n = 5 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \bar{x} = \dfrac{127.3}{5} \\[5ex] \bar{x} = 25.46 $
(3.) ACT A certain group consists of $5$ children, $3$ of whom are age $10$ and $2$ of whom are age $5$.
What is the mean age of the children in the group?

$ A.\:\: 5 \\[3ex] B.\:\: 7 \\[3ex] C.\:\: 7.5 \\[3ex] D.\:\: 8 \\[3ex] E.\:\: 10 \\[3ex] $

$ 3\:\:children\:\:aged\:\:10 = 10, 10, 10 \\[3ex] 2\:\:children\:\:aged\:\:5 = 5, 5 \\[3ex] x = Ages = 10, 10, 10, 5, 5 \\[3ex] \Sigma x = 10 + 10 + 10 + 5 + 5 \\[3ex] \Sigma x = 40 \\[3ex] n = 5 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \bar{x} = \dfrac{40}{5} \\[5ex] \bar{x} = 8 $
(4.) ACT To increase the mean of $4$ numbers by $3$, by how much would the sum of the $4$ numbers have to increase?

$ F.\:\: \dfrac{3}{4} \\[5ex] G.\:\: 1 \\[3ex] H.\:\: \dfrac{4}{3} \\[5ex] J.\:\: 7 \\[3ex] K.\:\: 12 \\[3ex] $

$ \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] n = 4 \\[3ex] Increase\:\:\bar{x}\:\:by\:\: 3 \rightarrow \bar{x} + 3 \\[3ex] \therefore \bar{x} + 3 = \dfrac{\Sigma x}{n} + 3 \\[5ex] \bar{x} + 3 = \dfrac{\Sigma x}{4} + 3 \\[5ex] \bar{x} + 3 = \dfrac{\Sigma x}{4} + \dfrac{12}{4} \\[5ex] \bar{x} + 3 = \dfrac{\Sigma x + 12}{4} \\[5ex] $ $\therefore \Sigma x$ should be increased by $12$
(5.) JAMB Thirty boys and $x$ girls sat for a test.
The mean of the boys' scores and that of the girls were respectively $6$ and $8$
Find $x$ if the total score was $468$

$ A.\:\: 38 \\[3ex] B.\:\: 24 \\[3ex] C.\:\: 36 \\[3ex] D.\:\: 22 \\[3ex] E.\:\: 41 \\[3ex] $

$ \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \Sigma x = \bar{x} * n \\[3ex] \underline{Boys} \\[3ex] n = 30 \\[3ex] \bar{x} = 6 \\[3ex] \Sigma x = 6(30) \\[3ex] \Sigma x = 180 \\[3ex] \underline{Girls} \\[3ex] n = x \\[3ex] \bar{x} = 8 \\[3ex] \Sigma x = 8(x) = 8x \\[3ex] \underline{Boys\:\:and\:\:Girls} \\[3ex] \Sigma x = 468 \\[3ex] 180 + 8x = 468 \\[3ex] 8x = 468 - 180 \\[3ex] 8x = 288 \\[3ex] x = \dfrac{288}{8} \\[5ex] x = 36 \\[3ex] $ There were $36$ girls
(6.) ACT What is the product of the mean and median of the first $6$ prime numbers?
(Note: $2$ is the first prime number.)

$ A.\:\: 27 \\[3ex] B.\:\: 37 \\[3ex] C.\:\: 39 \\[3ex] D.\:\: 41 \\[3ex] E.\:\: 42 \\[3ex] $

$ The\:\:first\:\:6\:\:prime\:\:numbers\:\:are\:\: 2, 3, 5, 7, 11, 13 \\[3ex] Mean = \dfrac{2 + 3 + 5 + 7 + 11 + 13}{6} = \dfrac{41}{6} \\[5ex] Median = \dfrac{5 + 7}{2} = \dfrac{12}{2} = 6 \\[5ex] Mean * Median = \dfrac{41}{6} * 6 = 41 $
(7.) At the end of the second semester, Rita earned an $A$ in the $4-credit$ Algebra course, a $B$ in the $3-credit$ English course, a $B$ in the $4-credit$ Biology course, an $A$ in the $4-credit$ Chemistry course, and a $C$ in the $1-credit$ Chinese Language course.
Using a $4-point$ grade system, $A$ is equivalent to $4$ points, $B$ is equivalent to $3$ points, $C$ is equivalent to $2$ points, $D$ is equivalent to $1$ point, and $F$ has no point.
Calculate Rita's grade point average ($GPA$) for the second semester.


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.) Based on Question $(7).$
Rita had a $GPA$ of approximately $3.58$ (actual GPA = 3.583333333) during the first semester when she took $12-credit$ hours.
Calculate her cumulative grade point average ($CGPA$) for the year (for both first and second semesters)


$ \underline{First\:\:Semester} \\[3ex] \Sigma CH = 12 \\[3ex] GPA = 3.583333333 \\[3ex] \Sigma GP = 3.583333333(12) = 43 \\[3ex] \underline{Second\:\:Semester} \\[3ex] \Sigma CH = 16 \\[3ex] \Sigma GP = 55 \\[3ex] \underline{First\:\:and\:\:Second\:\:Semesters} \\[3ex] \Sigma CH = 12 + 16 = 28 \\[3ex] \Sigma GP = 43 + 55 = 98 \\[3ex] CGPA = \dfrac{\Sigma GP}{\Sigma CH} \\[5ex] CGPA = \dfrac{98}{28} \\[5ex] CGPA = 3.5 \\[3ex] CGPA = 3.50 $
(9.) JAMB Use the table below to answer the questions.

Marks $1$ $2$ $3$ $4$ $5$
Frequencies $2$ $2$ $8$ $4$ $4$

How many students took the test?

$ A.\:\: 16 \\[3ex] B.\:\: 20 \\[3ex] C.\:\: 13 \\[3ex] D.\:\: 15 \\[3ex] $ Find the mean mark.

$ A.\:\: 3.1 \\[3ex] B.\:\: 3.0 \\[3ex] C.\:\: 3.3 \\[3ex] D.\:\: 3.2 \\[3ex] $

Marks, $x$ Frequencies, $f$ $f * x$
$1$ $2$ $2$
$2$ $2$ $4$
$3$ $8$ $24$
$4$ $4$ $16$
$5$ $4$ $20$
$\Sigma f = 20$ $\Sigma fx = 66$

$20$ students took the test

$ \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] \bar{x} = \dfrac{66}{20} = \dfrac{33}{10} \\[5ex] \bar{x} = 3.3 $
(10.) ACT For $20$ quiz scores in a typing class, the table below gives the frequency of the scores in each score interval.
Which score interval contains the median of the scores?

Score interval Frequency
$96-100$ $3$
$91-95$ $1$
$86-90$ $3$
$81-85$ $4$
$76-80$ $9$

$ A.\:\: 96-100 \\[3ex] B.\:\: 91-95 \\[3ex] C.\:\: 96-90 \\[3ex] D.\:\: 81-85 \\[3ex] E.\:\: 76-80 \\[3ex] $

$ \Sigma f = 3 + 1 + 3 + 4 + 9 = 20 \\[3ex] \dfrac{\Sigma f}{2} = \dfrac{20}{2} = 10 \\[5ex] Begin\:\:from\:\:the\:\:first\:\:class \\[3ex] Keep\:\:adding\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:to\:\:10 \\[3ex] 3 + 1 = 4 \\[3ex] 4 + 3 = 7 \\[3ex] 7 + 4 = 11...stop \\[3ex] $ The class that contains the median is $81-85$
(11.) WASCCE If the mean of the numbers $5, 8, x, 12, (x + 5)\:\:and\:\:10\:\:is\:\:10$, find $x$

$ A.\:\: 6 \\[3ex] B.\:\: 8 \\[3ex] C.\:\: 10 \\[3ex] D.\:\: 60 \\[3ex] $

$ \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.) A nurse recorded the weights of $2$ patients in a hospital.
She noticed that one of the weights she recorded was not clear.
She wanted to re-record but all the patients were discharged.
Assume the mean score of the weights is $82$ and the mean of the $19$ readable weights is $86$, what is the value of the unreadable weight?


Let the unreadable weight = $x$

$ For\:\:20\:\:patients \\[3ex] \bar{x} = 82 \\[3ex] n = 20 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \rightarrow \Sigma x = \bar{x} * n \\[3ex] \Sigma x = 82(20) = 1640 \\[3ex] For\:\:19\:\:patients \\[3ex] \bar{x} = 86 \\[3ex] n = 19 \\[3ex] \Sigma x = \bar{x} * n \\[3ex] \Sigma x = 86(19) = 1634 \\[3ex] \Sigma x_{19} + x = \Sigma x_{20} \\[3ex] 1634 + x = 1640 \\[3ex] x = 1640 - 1634 \\[3ex] x = 6 $
(13.) JAMB Find the mean of the data $7, -3, 4, -2, 5, -9, 4, 8, -6, 12$

$ A.\:\: 4 \\[1em] B.\:\: 3 \\[1em] C.\:\: 2 \\[1em] D.\:\: 1 \\[1em] $

$ x = 7, -3, 4, -2, 5, -9, 4, 8, -6, 12 \\[3ex] \Sigma x = 7 + -3 + 4 + -2 + 5 + -9 + 4 + 8 + -6 + 12 \\[3ex] \Sigma x = 7 - 3 + 4 - 2 + 5 - 9 + 4 + 8 - 6 + 12 \\[3ex] \Sigma x = 20 \\[3ex] n = 10 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \bar{x} = \dfrac{20}{10} \\[5ex] \bar{x} = 2 $
(14.) The retirement ages (in years) of nine employees of the town of Scrabble, West Virginia are:
$51, 61, 62, 57, 50, 67, 68, 58, 53$
Determine the mode.


Data Value Frequency
$51$ $1$
$61$ $1$
$62$ $1$
$57$ $1$
$50$ $1$
$67$ $1$
$68$ $1$
$58$ $1$
$53$ $1$

There is no mode.

(15.) CSEC (a) The information below represents the minimum temperatures, in $^\circ C$, recorded in Country A for the first $20$ days in a particular month.

$21$ $23$ $25$ $22$ $24$ $25$ $23$ $26$ $23$ $24$
$25$ $26$ $23$ $25$ $23$ $25$ $24$ $25$ $25$ $25$

(i) Complete the frequency table below, using the information above.

Temperature ($^\circ C$) Tally Frequency
$21$ I $1$
$22$ I $1$
$23$
$24$ III $3$
$25$
$26$

(ii) Determine the median temperature.

(iii) Calculate the mean temperature for the twenty-day period.


(i) The completed frequency table is:
Temperature, $x$ ($^\circ C$) Tally Frequency, $f$ $f * x$
$21$ I $1$ $21$
$22$ I $1$ $22$
$23$ IIII $5$ $115$
$24$ III $3$ $72$
$25$ IIII III $8$ $200$
$26$ II $2$ $52$
$\Sigma f = 20$ $\Sigma fx = 482$

$ \Sigma f = 1 + 1 + 5 + 3 + 8 + 2 = 20 \\[3ex] An\:\:even\:\:sample\:\:size\:\:means\:\:two\:\:middle\:\:numbers \\[3ex] \dfrac{\Sigma f}{2} = \dfrac{20}{2} = 10 \\[5ex] Begin\:\:from\:\:the\:\:first\:\:temperature \\[3ex] Keep\:\:adding\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:to\:\:10 \\[3ex] 1 + 1 = 2 \\[3ex] 2 + 5 = 7 \\[3ex] 7 + 3 = 10...stop \\[3ex] $ The temperature that contains that $10$ is $24$

$ Begin\:\:from\:\:the\:\:last\:\:temperature \\[3ex] Keep\:\:adding\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:to\:\:10 \\[3ex] 2 + 8 = 10...stop \\[3ex] $ The temperature that contains that $10$ is $25$
This implies that the two middle numbers are $24$ and $25$

$ \therefore \tilde{x} = \dfrac{24 + 25}{2} \\[5ex] \tilde{x} = \dfrac{49}{2} \\[5ex] \tilde{x} = 24.5 \\[3ex] $ (ii) The median temperature is $24.5^\circ C$

$ \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] \bar{x} = \dfrac{482}{20} \\[5ex] \bar{x} = 24.1 \\[3ex] $ (iii) The mean temperature is $24.1^\circ C$

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

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

Calculate the measures of central tendency of the data set.


To make it easier (as I did in the video),
Mark the $45-s$ with red circles
Mark the $48-s$ with red check marks
Mark the $53-s$ with red squares
Mark the $54-s$ with red triangles
Mark the $55-s$ with red hexagons
Mark the $58-s$ with white circles
Mark the $61-s$ with white triangles
Mark the $62-s$ with white squares
Mark the $63-s$ with white check marks
Mark the $64-s$ with white hexagons
Mark the $65-s$ with blue check marks
Mark the $66-s$ with blue circles
Mark the $68-s$ with blue squares
Mark the $70-s$ with blue triangles
Mark the $71-s$ if you wish
Mark the $75-s$ if you wish

Ages, $x$ Frequency, $f$ $f * x$
$45$ $3$ $135$
$48$ $1$ $48$
$53$ $2$ $106$
$54$ $1$ $54$
$55$ $1$ $55$
$58$ $1$ $58$
$61$ $2$ $122$
$62$ $1$ $62$
$63$ $3$ $189$
$64$ $1$ $64$
$65$ $2$ $130$
$66$ $2$ $132$
$68$ $1$ $68$
$70$ $2$ $140$
$71$ $1$ $71$
$75$ $1$ $75$
$\Sigma f = 25$ $\Sigma fx = 1509$

$ \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] \bar{x} = \dfrac{1509}{25} \\[5ex] \bar{x} = 60.36 \\[3ex] $ The mean age is $60.36$ years

$ \Sigma f = 25 \\[3ex] An\:\:odd\:\:sample\:\:size\:\:means\:\:only\:\:one\:\:middle\:\:number \\[3ex] \dfrac{\Sigma f}{2} = \dfrac{25}{2} = 12.5 \\[5ex] Begin\:\:from\:\:the\:\:first\:\:age \\[3ex] Keep\:\:adding\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:to\:\:12.5 \\[3ex] 3 + 1 = 4 \\[3ex] 4 + 2 = 6 \\[3ex] 6 + 1 = 7 \\[3ex] 7 + 1 = 8 \\[3ex] 8 + 1 = 9 \\[3ex] 9 + 2 = 11 \\[3ex] 11 + 1 = 12 \\[3ex] 12 + 3 = 15...stop \\[3ex] $ The age that contains that $12.5$ is $63$
The median age is $63$ years

Highest Frequency = $3$
The ages with the highest frequency are $45$ and $63$
Technically, this is a bimodal data ... it has two modes
However, looking at the values of the mean and median;
The "actual" modal age is $63$ years

$ x_{MR} = \dfrac{min + max}{2} \\[5ex] min = 45 \\[3ex] max = 75 \\[3ex] \rightarrow x_{MR} = \dfrac{45 + 75}{2} \\[5ex] x_{MR} = \dfrac{120}{2} \\[5ex] x_{MR} = 60 \\[3ex] $ The midrange age is $60$ years
(17.) CMAT Students in a commerce class took a test.
The average score of men is $70$ and of women is $83$.
The class average is $76$.
What is the ratio of men to women in the class?

$ 1.\:\: 8:7 \\[3ex] 2.\:\: 7:8 \\[3ex] 3.\:\: 7:6 \\[3ex] 4.\:\: 6:7 \\[3ex] $

$ \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \rightarrow \Sigma x = \bar{x} * n \\[3ex] \underline{Men} \\[3ex] n_m = ? \\[3ex] \bar{x}_m = 70 \\[3ex] \Sigma x_m = \bar{x}_m * n_m \\[3ex] \Sigma x_m = 70 * n_m \\[3ex] \underline{Women} \\[3ex] n_w = ? \\[3ex] \bar{x}_w = 83 \\[3ex] \Sigma x_w = \bar{x}_w * n_w \\[3ex] \Sigma x_w = 83 * n_w \\[3ex] \underline{Entire\:\:Class - Both\:\:Men\:\:and\:\:Women} \\[3ex] n_m + n_w = ? \\[3ex] \bar{x}_{mw} = 76 \\[3ex] \Sigma x_{mw} = \bar{x}_{mw} * (n_m + n_w) \\[3ex] \Sigma x_{mw} = 76 * (n_m + n_w) \\[3ex] $ The summation of the scores of the class (men and women) is equal to the sum of the summation of the scores of the men and the summation of the scores of the women.

$ \therefore 76(n_m + n_w) = 70n_m + 83n_w \\[3ex] 76n_m + 76n_w = 70n_m + 83n_w \\[3ex] 76n_m - 70n_m = 83n_w - 76n_w \\[3ex] 6n_m = 7n_w \\[3ex] \dfrac{n_m}{n_w} = \dfrac{7}{6} $
(18.) JAMB By how much is the mean of $30, 56, 31, 55, 43,\:and\:44$ less than the median?

$ A.\:\: 0.50 \\[3ex] B.\:\: 0.33 \\[3ex] C.\:\: 0.17 \\[3ex] D.\:\: 0.75 \\[3ex] $

The data set sorted in ascending order is $30, 31, 43, 44, 55, 56$

$ \bar{x} = \dfrac{30 + 31 + 43 + 44 + 55 + 56}{6} = \dfrac{259}{6} \\[5ex] \tilde{x} = \dfrac{43 + 44}{2} = \dfrac{87}{2} \\[5ex] Mean\:\:less\:\:than\:\:Median\:\:means\:\:Median - Mean \\[3ex] \tilde{x} - \bar{x} = \dfrac{87}{2} - \dfrac{259}{6} \\[5ex] = \dfrac{261}{6} - \dfrac{259}{6} \\[5ex] = \dfrac{261 - 259}{6} \\[5ex] = \dfrac{2}{6} \\[5ex] = \dfrac{1}{3} \approx 0.33 $
(19.) WASSCE-FM In a Physics examination, the mean mark of the first twelve students in a class is $60$, that of the next twenty students is $50$ and that of the remaining $y$ students is $x$. What is the mean mark for the whole class in the examination, in terms of $x$ and $y$?


$ \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \rightarrow \Sigma x = n * \bar{x} \\[3ex] \underline{First\:\:12\:\:students} \\[3ex] n = 12 \\[3ex] \bar{x} = 60 \\[3ex] \Sigma x = 12 * 60 = 720 \\[3ex] \underline{Next\:\:Twenty\:\:students} \\[3ex] n = 20 \\[3ex] \bar{x} = 50 \\[3ex] \Sigma x = 20 * 50 = 1000 \\[3ex] \underline{Remaining\:\:y\:\:students} \\[3ex] n = y \\[3ex] \bar{x} = x \\[3ex] \Sigma x = y * x = xy \\[3ex] \underline{Whole\:\:Class} \\[3ex] n = 12 + 20 + y = 32 + y \\[3ex] \Sigma x = 720 + 1000 + xy = 1720 + xy \\[3ex] \bar{x} = \dfrac{1720 + xy}{32 + y} \\[5ex] \bar{x} = \dfrac{xy + 1720}{y + 32} $
(20.) ACT The $5$ positive integers $x, x, x, y,\:\:and\:\: z$ have an average of $x$
Which of the following equations must be true?

$ A.\:\: y = -z \\[3ex] B.\:\: y = z \\[3ex] C.\:\: y + z = 0 \\[3ex] D.\:\: y + z = x \\[3ex] E.\:\: y + z = 2x \\[3ex] $

$ \Sigma x = x + x + x + y + z = 3x + y + z \\[3ex] n = 5 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \bar{x} = x \\[3ex] \rightarrow x = \dfrac{3x + y + z}{5} \\[5ex] Cross\:\:Multiply \\[3ex] 5x = 3x + y + z \\[3ex] 5x - 3x = y + z \\[3ex] 2x = y + z \\[3ex] y + z = 2x $




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(21.) WASSCE The marks scored by $50$ students in a Geography examination are as follows:

$60$ $50$ $40$ $67$ $53$ $73$ $37$ $55$ $62$ $43$
$44$ $69$ $39$ $32$ $45$ $58$ $48$ $67$ $39$ $51$
$46$ $59$ $40$ $52$ $61$ $48$ $23$ $60$ $59$ $47$
$65$ $58$ $74$ $47$ $40$ $59$ $68$ $51$ $50$ $50$
$71$ $51$ $26$ $36$ $38$ $70$ $46$ $40$ $51$ $42$

(a) Using class intervals $21-30, 31-40...$, prepare a frequency distribution table.

(b) Calculate the mean mark of the distribution.

(c) What percentage of the students scored more than $60$?


$ minimum = 23 \\[3ex] maximum = 74 \\[3ex] $ To make it easier,
Draw the table with the class intervals and tally first
Begin from the first column (not the first row because it is easier to deal with the data values with the way it is arranged in columns)
Place tallies as you read the data values from top to bottom
Then, draw the frequency column and add the frequencies
Check to make sure the sum of the frequencies is $50$

OR

Mark the $21-30$ with triangles
Mark the $31-40$ with squares
Mark the $41-50$ with circles
Mark the $51-60$ with $X's$
Mark the $61-70$ with check marks
Mark the $71-80$ if you wish

Marks, $x$ Tally Frequency, $f$ $x_{mid} = \dfrac{LCI + UCI}{2}$ $f * x_{mid}$
$21-30$ II $2$ $\dfrac{21 + 30}{2} = \dfrac{51}{2} = 25.5$ $51$
$31-40$ IIII IIII $10$ $\dfrac{31 + 40}{2} = \dfrac{71}{2} = 35.5$ $355$
$41-50$ IIII IIII III $12$ $\dfrac{41 + 50}{2} = \dfrac{91}{2} = 45.5$ $546$
$51-60$ IIII IIII IIII $15$ $\dfrac{51 + 60}{2} = \dfrac{111}{2} = 55.5$ $832.5$
$61-70$ IIII III $8$ $\dfrac{61 + 70}{2} = \dfrac{131}{2} = 65.5$ $524$
$71-80$ III $3$ $\dfrac{71 + 80}{2} = \dfrac{151}{2} = 75.5$ $226.5$
$\Sigma f = 50$ $\Sigma fx_{mid} = 2535$

$ (b) \\[1em] \bar{x} = \dfrac{\Sigma fx_{mid}}{\Sigma f} \\[5ex] \bar{x} = \dfrac{2535}{50} \\[5ex] \bar{x} = 50.7 \\[3ex] (c) \\[1em] More\:\:than\:\:60\:\:means\:\: 61-70\:\:and\:\:71-80 \\[3ex] Number\:\:of\:\:students\:\:who\:\:scored\:\:more\:\:than\:\:60 = 8 + 3 = 11 \\[3ex] Total\:\:number\:\:of\:\:students = \Sigma f = 50 \\[3ex] \therefore Percentage\:\:of\:\:students\:\:who\:\:scored\:\:more\:\:than\:\:60 = \dfrac{11}{50} * 100 \\[5ex] = 11 * 2 \\[3ex] = 22\% $

(22.) CSEC A class of $30$ students counted the number of books in their bags on a certain day. The number of books in EACH bag is shown below.

$5$ $4$ $6$ $3$ $2$ $1$ $7$ $4$ $5$ $3$
$6$ $5$ $4$ $3$ $7$ $6$ $2$ $5$ $4$ $5$
$5$ $7$ $5$ $4$ $3$ $2$ $1$ $6$ $3$ $4$

(a) Copy and complete the frequency table for the data shown above.

Number of Books $(x)$ Tally Frequency $(f)$ $f * x$
$1$ II $2$ $2$
$2$ III $3$ $6$
$3$ $...$ $...$
$4$ $...$ $...$
$5$ $...$ $...$
$6$ $...$ $...$
$7$ $...$ $...$

(b) State the modal number of books in the bags of the sample of students.

(c) Using the table in (a) above, or otherwise, calculate
(i) the TOTAL number of books
(ii) the mean number of books per bag.

(d) Determine the probability that a student chosen at random has LESS THAN $4$ books in his/her bag.


To make it easier,
Draw the table with the class intervals and tally first
Just begin with a clean table.. with no prior entries.
Begin from the first column (not the first row because it is easier to deal with the data values with the way it is arranged in columns)
Place tallies as you read the data values from top to bottom
Then, draw the frequency column and add the frequencies
Check to make sure the sum of the frequencies is $30$

Number of Books $(x)$ Tally Frequency $(f)$ $f * x$
$1$ II $2$ $2$
$2$ III $3$ $6$
$3$ IIII $5$ $15$
$4$ IIII I $6$ $24$
$5$ IIII II $7$ $35$
$6$ IIII $4$ $24$
$7$ III $3$ $21$
$\Sigma f = 30$ $\Sigma fx = 127$

$ (b)\:\: \Highest\:\:Frequency = 7 \\[3ex] Number\:\:with\:\:the\:\:highest\:\:frequency = 5 \\[3ex] \therefore the\:\:modal\:\:number\:\:of\:\:books = 5 \:\:books\:\:bag \\[3ex] (c)\:\:(i)\:\: Total\:\:number\:\:of\:\:books = \Sigma fx = 127 \:\:books \\[3ex] (ii)\:\: \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] \bar{x} = \dfrac{127}{30} \\[3ex] \bar{x} = 4.2333 \\[3ex] (d)\:\:Let\:\:E = event\:\:of\:\:less\:\:than\:\:4\:\:books \\[3ex] n(E) = 5 + 3 + 2 = 10 \\[3ex] n(S) = \Sigma f = 30 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{10}{30} \\[5ex] P(E) = \dfrac{1}{3} $
(23.) ACT The $5$ positive integers $a, a, a, b,\:\:and\:\: c$ have an average of $a$.
What is the value of $\dfrac{b + c}{2}$

$ F.\:\: \dfrac{a}{3} \\[5ex] G.\:\: \dfrac{a}{2} \\[5ex] H.\:\: a \\[3ex] J.\:\: 2a \\[3ex] K.\:\: 3a \\[3ex] $

$ \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \Sigma x = a + a + a + b + c = 3a + b + c \\[3ex] n = 5 \\[3ex] \bar{x} = a \\[3ex] \rightarrow a = \dfrac{3a + b + c}{5} \\[5ex] Cross\:\:Multiply \\[3ex] 5a = 3a + b + c \\[3ex] 5a - 3a = b + c \\[3ex] 2a = b + c \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:2 \\[3ex] a = \dfrac{b + c}{2} \\[5ex] \dfrac{b + c}{2} = a $
(24.) ACT In a town of $500$ people, the $300$ males have an average age of $45$ and the $200$ females have an average age of $35$
To the nearest year, what is the average age of the town's entire population?

$ A.\:\: 40 \\[3ex] B.\:\: 41 \\[3ex] C.\:\; 42 \\[3ex] D.\:\: 43 \\[3ex] E.\:\: 44 \\[3ex] $

$ \underline{Males} \\[3ex] n = 300 \\[3ex] \bar{x} = 45 \\[3ex] \Sigma x = \bar{x} * n \\[3ex] \Sigma x = 45 * 300 \\[3ex] \Sigma x = 13500 \\[3ex] \underline{Females} \\[3ex] n = 200 \\[3ex] \bar{x} = 35 \\[3ex] \Sigma x = \bar{x} * n \\[3ex] \Sigma x = 35 * 200 \\[3ex] \Sigma x = 7000 \\[3ex] \underline{Entire\:\:Town} \\[3ex] n = 500 \\[3ex] \Sigma x = \Sigma x (Males) + \Sigma x (Females) \\[3ex] \Sigma x = 13500 + 7000 \\[3ex] \Sigma x = 20500 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \bar{x} = \dfrac{20500}{500} \\[5ex] \bar{x} = 41 \\[3ex] $ The average age of the town's entire population is $41$ years.
(25.) The assessments in Mr. C's online Precalculus class are $15$ discussion board assignments which includes $2$ projects, $26$ MML (My Math Lab) assignments, $3$ tests, and a comprehensive final exam.
The final grade of his students are computed using the Weighted Average Method.
He rounds the final grades of his students to the nearest integer only one time.
The DB assignments are worth $15\%$
The MML assignments constitute $60\%$
The tests are $15\%$
The final exam is $10\%$
Hannah made a combined score of: $80\%$ on the DB assignments and $90\%$ on the MML assignments.
Her test scores were $96, 92, 84$ percent respectively.
She made an $80\%$ on the final exam.
Calculate her final grade.


$ 15\:\:DB\:\:Assignments\:\:of\:\:80\% \:\:weighted\:\: 15\% = 80(15) = 1200 \\[3ex] 26\:\:MML\:\:Assignments\:\:of\:\:90\% \:\:weighted\:\: 60\% = 90(60) = 5400 \\[3ex] Test\:\:1\:\:of\:\:96\% \:\:weighted\:\: 5\% = 96(5) = 480 \\[3ex] Test\:\:2\:\:of\:\:92\% \:\:weighted\:\: 5\% = 92(5) = 460 \\[3ex] Test\:\:3\:\:of\:\:84\% \:\:weighted\:\: 5\% = 84(5) = 420 \\[3ex] Final\:\:Exam\:\:of\:\:80\% \:\:weighted\:\: 10\% = 80(10) = 800 \\[3ex] Final\:\:Grade = \dfrac{1200 + 5400 + 480 + 460 + 420 + 800}{15 + 60 + 5 + 5 + 5 + 10} = \dfrac{8760}{100} \\[5ex] Final\:\:Grade = 87.6\% \approx 88\% $
(26.) JAMB The mean of seven numbers is $96$.
If an eighth number is added, the mean becomes $112$.
Find the eighth number.

$ A.\:\: 126 \\[3ex] B.\:\: 180 \\[3ex] C.\:\: 216 \\[3ex] D.\:\: 224 \\[3ex] $

$ \Sigma x_7 = sum\:\:of\:\:7\:\:subjects \\[3ex] \Sigma x_8 = sum\:\:of\:\:8\:\:subjects \\[3ex] For\:\:7\:\:numbers \\[3ex] \bar{x} = 96 \\[3ex] n = 7 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \rightarrow \Sigma x = \bar{x} * n \\[3ex] \Sigma x_7 = 96 * 7 = 672 \\[3ex] For\:\:8\:\:numbers \\[3ex] \bar{x} = 112 \\[3ex] n = 8 \\[3ex] \Sigma x_8 = 112 * 8 = 896 \\[3ex] Eighth\:\:number = \Sigma x_8 - \Sigma x_7 = 896 - 672 \\[3ex] \therefore Eighth\:\:number = 224 $
(27.) The assessments in Mr. C's online Prealgebra class are $32$ MOM (My Open Math) assignments, $3$ tests, and a comprehensive final exam.
The final grade of his students are computed using the Weighted Average Method.
He rounds the final grades of his students to the nearest integer only one time.
The MOM assignments are worth $60\%$
Each test is $10\%$
The final exam is also $10\%$
Joel made a combined score of: $91.38\%$ on the MOM assignments.
His test scores are $90, 65, 40$ percent respectively.
He wants to make an A in the class. Can he make an A?
If he misses an A, he does not want to settle for anything less than a B. Can he make a B?
Calculate the score which he must make in the final exam to earn an A? a B?
If he cannot make an A or a B (but still has the opportunity to improve his grades), what would you advise him to do?


$ 32\:\:MOM\:\:Assignments\:\:of\:\:91.38\% \:\:weighted\:\: 60\% = 91.38(60) = 5482.8 \\[3ex] Test\:\:1\:\:of\:\:90\% \:\:weighted\:\: 10\% = 90(10) = 900 \\[3ex] Test\:\:2\:\:of\:\:65\% \:\:weighted\:\: 10\% = 65(10) = 650 \\[3ex] Test\:\:3\:\:of\:\:40\% \:\:weighted\:\: 10\% = 40(10) = 400 \\[3ex] Let\:\:the\:\:final\:\:exam\:\:score = p \\[3ex] Final\:\:Exam\:\:of\:\:p\% \:\:weighted\:\: 10\% = p(10) = 10p \\[3ex] Final\:\:Grade = \dfrac{5482.8 + 900 + 650 + 400 + 10p}{60 + 10 + 10 + 10 + 10} = \dfrac{7432.8 + 10p}{100} \\[5ex] Grade\:\:of\:\:A\:\:is \ge 90\% \\[3ex] \implies \dfrac{7432.8 + 10p}{100} \ge 90 \\[5ex] LCD = 100 \\[3ex] Multiply\:\:both\:\:sides\:\:by \:\: 100 \\[3ex] 100 * \dfrac{7432.8 + 10p}{100} \ge 100 * 90 \\[5ex] 7432.8 + 10p \ge 9000 \\[3ex] 10p \ge 9000 - 7432.8 \\[3ex] 10p \ge 1567.2 \\[3ex] p \ge \dfrac{1567.2}{10} \\[5ex] p \ge 156.72\% \\[3ex] Maximum = 100\% \\[3ex] Because\:\: 156.72\% \gt 100\%,\:\:he\:\:cannot\:\:make\:\:an\:\:A \\[3ex] Grade\:\:of\:\:B\:\:is\:\:between\:\: 80\%(inclusive)\:\:and\:\:89.49999999999999\%(inclusive) \\[3ex] \implies \dfrac{7432.8 + 10p}{100} \ge 80 \\[5ex] LCD = 100 \\[3ex] Multiply\:\:both\:\:sides\:\:by \:\: 100 \\[3ex] 100 * \dfrac{7432.8 + 10p}{100} \ge 100 * 80 \\[5ex] 7432.8 + 10p \ge 8000 \\[3ex] 10p \ge 8000 - 7432.8 \\[3ex] 10p \ge 567.2 \\[3ex] p \ge \dfrac{567.2}{10} \\[5ex] p \ge 56.72\% \\[3ex] Joel\:\:can\:\:still\:\:make\:\:a\:\:B \\[3ex] $ He needs to make at least $56.72\%$ ($56.72\%$ or better) on the final exam in order to make a $B$
(28.) ACT The mean of $5$ integers is $52$
The median of these $5$ integers is $82$
Three of the integers are $0, 12$, and $82$
Which of the following could be one of the other integers?

$ F.\:\: 52 \\[3ex] G.\:\: 66 \\[3ex] H.\:\: 84 \\[3ex] J.\:\: 86 \\[3ex] K.\:\: 105 \\[3ex] $

$ n = 5 \\[3ex] \bar{x} = 52 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \Sigma x = \bar{x} * n \\[3ex] \Sigma x = 52(5) \\[3ex] \Sigma x = 260 \\[3ex] \tilde{x} = 82...means\:\:the\:\:third\:\:number\:\:for\:\:sample\:\:of\:\:five \\[3ex] \implies 0, 12, 82, a, c \\[3ex] \rightarrow 0 + 12 + 82 + a + c = 260 \\[3ex] 94 + a + c = 260 \\[3ex] a + c = 260 - 94 \\[3ex] a + c = 166...Condition\:\:1 \\[3ex] fourth\:\:number\:\:\ge 82...Condition\:\:2 \\[3ex] Options\:\:F\:\:and\:\:G\:\:are\:\:eliminated \\[3ex] Assume\:\:fourth\:\:number = 82...the\:\:minimum \\[3ex] 82 + 84 = 166 = 166 ...possible \\[3ex] 82 + 86 = 168 \gt 166 ...not\:\:possible \\[3ex] 82 + 105 = 187 \gt 166 ...not\:\:possible \\[3ex] Correct\:\:Option = 84 = H $
(29.) WASSCE In a test, if a student had scored $80$ marks in one of the subjects, his average mark in $8$ subjects would be $62$.
If he had scored $64$ marks in that same subject with the scores in the remaining $7$ subjects unchanged, the average mark would be $m$
Find the value of $m$.


$ \Sigma x_8 = sum\:\:of\:\:8\:\:subjects \\[3ex] \Sigma x_7 = sum\:\:of\:\:7\:\:subjects \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \underline{First\:\:Scenario} \\[3ex] n = 8 \\[3ex] \Sigma x = \Sigma_8 = \Sigma x_7 + 80 \\[3ex] \bar{x} = 62 \\[3ex] \rightarrow 62 = \dfrac{\Sigma x_7 + 80}{8} \\[5ex] \Sigma x_7 + 80 = 62(8) \\[3ex] \Sigma x_7 + 80 = 496 \\[3ex] \Sigma x_7 = 496 - 80 = 416 \\[3ex] \underline{Second\:\:Scenario} \\[3ex] n = 8 \\[3ex] \Sigma x = \Sigma_8 = \Sigma x_7 + 64 = 416 + 64 = 480 \\[3ex] \bar{x} = m \\[3ex] \therefore m = \bar{x} = \dfrac{\Sigma x}{n} = \dfrac{480}{8} = 60 \\[5ex] m = 60 $
(30.) JAMB Find the mean of $t + 2$, $2t - 4$, $3t + 2$, and $2t$.

$ A.\:\: 2t \\[1em] B.\:\: 2t + 1 \\[1em] C.\:\: t \\[1em] D.\:\: t + 1 \\[1em] $

$ x = t + 2, 2t - 4, 3t + 2, 2t \\[1em] n = 4 \\[1em] \Sigma x = (t + 2) + (2t - 4) + (3t + 2) + 2t \\[1em] \Sigma x = t + 2 + 2t - 4 + 3t + 2 + 2t \\[1em] \Sigma x = 8t \\[1em] \bar{x} = \dfrac{\Sigma x}{n} \\[2em] \bar{x} = \dfrac{8t}{4} \\[2em] \bar{x} = 2t $
(31.) ACT Ms. Simons made a bar graph of the $20$ scores on the last math test, as shown below.
Which of the following statements about the mean of the $20$ scores is true?
Bar Graph

F. The mean is less than $75$
G. The mean is $75$
H. The mean is between $75$ and $85$
J. The mean is $85$
K. The mean is greater than $85$


$ Based\:\:on\:\:the\:\:bar\:\:graph \\[1em] 1\:\:student\:\:scored\:\:60 \\[1em] 1\:\:student\:\:scored\:\:65 \\[1em] 1\:\:student\:\:scored\:\:70 \\[1em] 1\:\:student\:\:scored\:\:75 \\[1em] 2\:\:students\:\:scored\:\:80 \\[1em] 3\:\:students\:\:scored\:\:85 \\[1em] 2\:\:students\:\:scored\:\:90 \\[1em] 4\:\:students\:\:scored\:\:95 \\[1em] 5\:\:students\:\:scored\:\:100 \\[1em] Number\:\:of\:\:students = 1 + 1 + 1 + 1 + 2 + 3 + 2 + 4 + 5 = 20\:\:students...to\:\:confirm\:\:20\:\:scores \\[1em] $
Scores, $x$ Frequencies, $f$ $f * x$
$60$ $1$ $60$
$65$ $1$ $65$
$70$ $1$ $70$
$75$ $1$ $75$
$80$ $2$ $160$
$85$ $3$ $255$
$90$ $2$ $180$
$95$ $4$ $380$
$100$ $5$ $500$
$\Sigma f = 20$ $\Sigma fx = 1745$

$ \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] \bar{x} = \dfrac{1745}{20} \\[5ex] \bar{x} = 87.25 \\[1em] 87.25 \gt 85 $
(32.) ACT Set A and Set B each consist of $5$ distinct numbers.
The $2$ sets contain identical numbers with the exception of the number with the least value in each set.
The number with the least value in Set B is greater than the number with the least value in Set A.
The value of which of the following measures must be greater for Set B than for Set A?

A. Mean only
B. Median only
C. Mode only
D. Mean and median only
E. Mean, median, and mode


Let us write the two sets and ensure that the elements of the sets are listed in ascending order (from least to greatest).
We write these sets based on the information we were given.
Assume that:

$ A = \{1, 3, 4, 5, 6\} \\[1em] B = \{2, 3, 4, 5, 6\}...2 \gt 1 \\[1em] Median\:\:is\:\:the\:\:same = 4 \\[1em] Mode\:\:is\:\:the\:\:same...no\:\:mode \\[1em] Mean\:\:is\:\:different \\[1em] Mean\:\:of\:\:Set\:B \gt Mean\:\:of\:\:Set\: A \\[1em] How? \\[1em] \bar{x}\:\:of\:\:Set\:B = \dfrac{2 + 3 + 4 + 5 + 6}{5} = \dfrac{20}{5} = 4 \\[2em] \bar{x}\:\:of\:\:Set\:A = \dfrac{1 + 3 + 4 + 5 + 6}{5} = \dfrac{19}{5} \lt 4 $


Use the table below to answer the questions $33$ and $34$

JAMB
Score $4$ $7$ $8$ $11$ $13$ $8$
Frequency $3$ $5$ $2$ $7$ $2$ $1$


(33.) JAMB The mean score is

$ A.\:\: 7.0 \\[1em] B.\:\: 8.7 \\[1em] C.\:\: 9.5 \\[1em] D.\:\: 11.0 \\[1em] $

Score, $x$ Frequency, $f$ $f * x$
$4$ $3$ $12$
$7$ $5$ $35$
$8$ $2$ $16$
$11$ $7$ $77$
$13$ $2$ $26$
$8$ $1$ $8$
$\Sigma f = 20$ $\Sigma fx = 174$

$ \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] \bar{x} = \dfrac{174}{20} = \dfrac{87}{10} \\[5ex] \bar{x} = 8.7 $
(34.) JAMB Find the square of the mode.

$ A.\:\: 49 \\[1em] B.\:\: 121 \\[1em] C.\:\: 25 \\[1em] D.\:\: 64 \\[1em] $

The mode is the score with the highest frequency.

Highest Frequency $= 7$

Score with $7 = 11$

Mode $= 11$

Square of the mode $= 11^2 = 121$
(35.) WASSCE If the mean of $m$, $n$, $s$, $p$, and $q$ is $12$, calculate the mean of $(m + 4)$, $(n - 3)$, $(s + 6)$, $(p - 2)$, and $(q + 8)$


$ For:\:\:m, n, s, p, q \\[1em] \bar{x} = \dfrac{m + n + s + p + q}{5} = 12 \\[2em] For:\:\:(m + 4), (n - 3), (s + 6), (p - 2), (q + 8) \\[1em] \bar{x} = \dfrac{(m + 4) + (n - 3) + (s + 6) + (p - 2) + (q + 8)}{5} \\[2em] = \dfrac{m + 4 + n - 3 + s + 6 + p - 2 + q + 8}{5} \\[2em] = \dfrac{m + n + s + p + q + 13}{5} \\[2em] = \dfrac{m + n + s + p + q}{5} + \dfrac{13}{5} \\[2em] = 12 + \dfrac{13}{5} \\[2em] = \dfrac{60}{5} + \dfrac{13}{5} \\[2em] = \dfrac{60 + 13}{5} \\[2em] = \dfrac{73}{5} \\[2em] = 14\dfrac{3}{5} $
(36.) WASSCE The arithmetic mean of $x$, $y$, and $z$ is $6$ while that of $x$, $y$, $z$, $t$, $u$, $v$, and $w$ is $9$.
Calculate the arithmetic mean of $t$, $u$, $v$, and $w$


$ \bar{x} = \dfrac{\Sigma x}{n} \\[2em] For\:\:x, y, z \\[1em] n = 3 \\[1em] \Sigma x = x + y + z \\[1em] \bar{x} = 6 \\[1em] \implies 6 = \dfrac{x + y + z}{3} \\[2em] x + y + z = 6(3) \\[1em] x + y + z = 18...eqn.(1) \\[1em] For\:\: x, y, z, t, u, v, w \\[1em] n = 7 \\[1em] \Sigma x = x + y + z + t + u + v + w \\[1em] \bar{x} = 9 \\[1em] \implies 9 = \dfrac{x + y + z + t + u + v + w}{7} \\[2em] x + y + z + t + u + v + w = 9(7) \\[1em] x + y + z + t + u + v + w = 63...eqn.(2) \\[1em] Substitute\:\:eqn.(1)\:\:into\:\:eqn.(2) \\[1em] 18 + t + u + v + w = 63 \\[1em] t + u + v + w = 63 - 18 \\[1em] t + u + v + w = 45 \\[1em] For\:\:t, u, v, w \\[1em] n = 4 \\[1em] \Sigma x = 45 \\[1em] \bar{x} = \dfrac{45}{4} \\[2em] \bar{x} = 11.25 $
(37.) JAMB
Marks $2$ $3$ $4$ $5$ $6$ $7$ $8$
No. of students $3$ $1$ $5$ $2$ $4$ $2$ $3$

From the table above, if the pass mark is $5$, how many students failed the test?

$ A.\:\: 6 \\[3ex] B.\:\: 2 \\[3ex] C.\:\: 9 \\[3ex] D.\:\: 7 \\[3ex] $

$ 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 $
(38.) ACT The average weight of $10$ boys is $77.0$ pounds.
If the youngest boy is excluded, the avearge weight of the $9$ remaining boys is $78.0$ pounds.
What is the weight, in pounds, of the youngest boy?

$ A.\:\: 62 \\[3ex] B.\:\: 68 \\[3ex] C.\:\: 70 \\[3ex] D.\:\: 78 \\[3ex] E.\:\: 87 \\[3ex] $

$ \bar{x} = \dfrac{\Sigma x}{n} \\[2em] \Sigma x = n * \bar{x} \\[1em] n = 10 \\[1em] \bar{x}_{10} = 77 \\[1em] \Sigma x_{10} = 10 * 77 = 770 \\[1em] n = 9 \\[1em] \bar{x}_9 = 78 \\[1em] \Sigma x_9 = 9 * 78 = 702 \\[1em] Let\:\:the\:\:weight\:\:of\:\:the\:\:youngest\:\:boy = p \\[1em] \Sigma x_9 + p = \Sigma x_{10} \\[1em] 702 + p = 770 \\[1em] p = 770 - 702 \\[1em] p = 68 \\[1em] $ The youngest boy's weight is $68$ pounds.
(39.) JAMB The mean of a set of six numbers is $60$
If the mean of the first five is $50$, find the sixth number in the set.

$ A.\:\: 95 \\[1em] B.\:\: 100 \\[1em] C.\:\: 110 \\[1em] D.\:\: 105 \\[1em] $

$ \Sigma x_6 = sum\:\:of\:\:six\:\:numbers \\[3ex] \Sigma x_5 = sum\:\:of\:\:first\:\:five\:\:numbers \\[3ex] For\:\:6\:\:numbers \\[3ex] \bar{x} = 60 \\[3ex] n = 6 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \rightarrow \Sigma x = \bar{x} * n \\[3ex] \Sigma x_6 = 60 * 6 = 360 \\[3ex] For\:\:5\:\:numbers \\[3ex] \bar{x} = 50 \\[3ex] n = 5 \\[3ex] \Sigma x_5 = 50 * 5 = 250 \\[3ex] Sixth\:\:number = \Sigma x_6 - \Sigma x_5 = 360 - 250 \\[3ex] \therefore Sixth\:\:number = 110 $
(40.) JAMB By how much is the mean of $30, 56, 31, 55, 43$ and $44$ less than the median?

$ A.\:\: 0.50 \\[1em] B.\:\: 0.33 \\[1em] C.\:\: 0.17 \\[1em] D.\:\: 0.75 \\[1em] $

$ x = 30, 56, 31, 55, 43, 44 \\[1em] Sort\:\:in:\:ascending\:\:order \\[1em] x = 30, 31, 43, 44, 55, 56 \\[1em] \Sigma x = 30 + 31 + 43 + 44 + 55 + 56 \\[1em] \Sigma x = 259 \\[1em] n = 6 \\[1em] \bar{x} = \dfrac{\Sigma x}{n} \\[2em] \bar{x} = \dfrac{259}{6} \\[2em] \tilde{x} = \dfrac{44 + 55}{2} \\[2em] \tilde{x} = \dfrac{99}{2} \\[2em] Median - Mean \\[1em] = \tilde{x} - \bar{x} \\[1em] = \dfrac{99}{2} - \dfrac{296}{6} \\[2em] = \dfrac{297}{6} - \dfrac{296}{6} \\[2em] = \dfrac{297 - 296}{6} \\[2em] = \dfrac{1}{6} \\[2em] = 0.1666666667 \approx 0.17 $




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(41.) JAMB The mean age of a group of students is $15$ years.
When the age of a teacher, $45$ years old, is added to the ages of the students, the mean of their ages becomes $18$ years.
Find the number of students in the group.

$ A.\:\: 42 \\[1em] B.\:\: 15 \\[1em] C.\:\: 7 \\[1em] D.\:\: 9 \\[1em] $

$ \bar{x} = \dfrac{\Sigma x}{n} \\[2em] \Sigma x = \bar{x} * n \\[1em] Let\:\:the\:\:number\:\:of\:\:students = n \\[1em] For\:\:n \\[1em] \bar{x} = 15 \\[1em] \Sigma x = 15 * n = 15n \\[1em] Teacher\:\:is\:\:added \\[1em] Teacher\:\:is\:\:45\:\:years \\[1em] Sample\:\:size = n + 1 \\[1em] For\:\: n + 1 \\[1em] \bar{x} = 18 \\[1em] \Sigma x = 15n + 45 \\[1em] Also:\:\: \Sigma x = 18 * (n + 1) = 18(n + 1) \\[1em] \implies 15n + 45 = 18(n + 1) \\[1em] 15n + 45 = 18n + 18 \\[1em] 45 - 18 = 18n - 15n \\[1em] 27 = 3n \\[1em] 3n = 27 \\[1em] n = \dfrac{27}{3} \\[2em] n = 9 \\[1em] $ There are $9$ students in the group.

$ \underline{Check} \\[1em] Only\:\:the\:\:students \\[1em] n = 9 \\[1em] \bar{x} = 15 \\[1em] \Sigma x = 15(9) = 135 \\[1em] Students\:\:and\:\:Teacher \\[1em] n = 10 \\[1em] \Sigma x = 135 + 45 = 180 \\[1em] \bar{x} = \dfrac{180}{10} \\[2em] \bar{x} = 18 $
(42.) ACT The monthly fees for single rooms at $5$ colleges are $\$370$, $\$310$, $\$380$, $\$340$, and $\$310$, respectively.
What is the mean of these monthly fees?

$ A.\:\: \$310 \\[1em] B.\:\: \$340 \\[1em] C.\:\: \$342 \\[1em] D.\:\: \$350 \\[1em] E.\:\: \$380 \\[1em] $

$ x = 370, 310, 380, 340, 310 \\[1em] \Sigma x = 370 + 310 + 380 + 340 + 310 \\[1em] \Sigma x = 1710 \\[1em] n = 5 \\[1em] \bar{x} = \dfrac{\Sigma x}{n} \\[2em] \bar{x} = \dfrac{1710}{5} \\[2em] \bar{x} = \$342 $
(43.) JAMB The mean of seven numbers is $10$
If six of the numbers are $2, 4, 8, 14, 16$ and $18$, find the mode.

$ A.\:\: 8 \\[1em] B.\:\: 14 \\[1em] C.\:\: 2 \\[1em] D.\:\: 6 \\[1em] $

We need to find the seventh number.
Then, we can find the mode.

$ \bar{x} = \dfrac{\Sigma x}{n} \\[2em] Let\:\:the\:\:seventh\:\:number = p \\[1em] x = 2, 4, 8, 14, 16, 18, p \\[1em] n = 7 \\[1em] \Sigma x = 2 + 4 + 8 + 14 + 16 + 18 + p \\[1em] \Sigma x = 62 + p \\[1em] \bar{x} = 10 \\[1em] \implies 10 = \dfrac{62 + p}{7} \\[2em] 62 + p = 10(7) \\[1em] 62 + p = 70 \\[1em] p = 70 - 62 \\[1em] p = 8 \\[1em] \therefore x = 2, 4, 8, 14, 16, 18, 8 \\[1em] Highest\:\:frequency = 2...because\:\:8\:\:occured\:\:two\:\:times \\[1em] Number\:\:with\:the\:\:highest\:\:frequency = 8 \\[1em] Mode = 8 $
(44.) JAMB Find the median of $2, 3, 7, 3, 4, 5, 8, 9, 9, 4, 5, 3, 4, 2, 4$ and $5$

$ A.\:\: 9 \\[1em] B.\:\: 8 \\[1em] C.\:\: 7 \\[1em] D.\:\: 4 \\[1em] $

$ x = 2, 3, 7, 3, 4, 5, 8, 9, 9, 4, 5, 3, 4, 2, 4, 5 \\[1em] Sort\:\:in:\:ascending\:\:order \\[1em] x = 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 7, 8, 9, 9 \\[1em] \tilde{x} = \dfrac{4 + 4}{2} \\[2em] \tilde{x} = \dfrac{8}{2} \\[2em] \tilde{x} = 2 $
(45.) ACT The table below shows the number of cars Jing sold each month last year.
What is the median of the data in the table?

Month Number of cars sold
January $25$
February $15$
March $22$
April $19$
May $16$
June $13$
July $19$
August $25$
September $26$
October $27$
November $28$
December $29$

$ F.\:\: 13 \\[1em] G.\:\: 16 \\[1em] H.\:\: 19 \\[1em] J.\:\: 20.5 \\[1em] K.\:\: 23.5 \\[1em] $

$ Arrange\:\:the\:\:data\:\:in\:\:ascending\:\:order \\[1em] 13, 15, 16, 19, 19, 22, 25, 25, 26, 27, 28, 29 \\[1em] Check\:\:to\:\:make\:\:sure\:\: n = 12 \\[1em] \tilde{x} = \dfrac{22 + 25}{2} \\[2em] \tilde{x} = \dfrac{47}{2} \\[2em] \tilde{x} = 23.5 $
(46.) JAMB
Age $20$ $25$ $30$ $35$ $40$ $45$
No. of people $3$ $5$ $1$ $1$ $2$ $3$

Calculate the median age of the frequency distribution in the table above.

$ A.\:\: 30 \\[1em] B.\:\: 35 \\[1em] C.\:\: 20 \\[1em] D.\:\: 25 \\[1em] $

We can solve this question in at least two ways.
Use any method you prefer.
However, because this is a JAMB question, the second method is recommended.

$ \underline{First\:\:Method:\:\: List\:\:the\:\:data\:\:values} \\[1em] 3\:\:people\:\:are\:\:20\:\:years \implies 20, 20, 20 \\[1em] 5\:\:people\:\:are\:\:25\:\:years \implies 25, 25, 25, 25, 25 \\[1em] 1\:\:person\:\:is\:\:30\:\:years \implies 30 \\[1em] 1\:\:person\:\:is\:\:35\:\:years \implies 35 \\[1em] 2\:\:people\:\:are\:\:40\:\:years \implies 40, 40 \\[1em] 3\:\:people\:\:are\:\:45\:\:years \implies 45, 45, 45 \\[1em] The\:\:dataset\:\:is \\[1em] 20, 20, 20, 25, 25, 25, 25, 25, 30, 35, 40, 40, 45, 45, 45 \\[1em] Median\:\:age\:\:is\:\:the\:\:middle\:\:number \\[1em] \tilde{x} = 25\:years \\[2em] \underline{Second\:\:Method} \\[1em] Age\:\:data\:\:is\:\:also\:\:sorted\:\:in\:\:ascending\:\:order \\[1em] \Sigma F = 3 + 5 + 1 + 1 + 2 + 3 = 15 \\[1em] \dfrac{\Sigma F}{2} = \dfrac{15}{2} = 7.5 \\[2em] Begin\:\:from\:\:the\:\:first\:\:age \\[1em] Keep\:\:adding\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:7.5 \\[1em] 3 \\[1em] 3 + 5 = 8 \\[1em] 8 \:\:contains\:\: 7.5 \\[1em] STOP \\[1em] The\:\:age\:\:that\:\:has\:\:the\:\:median\:\:is\:\:25\:years \\[1em] \tilde{x} = 25\:years $
(47.) ACT The difference (larger minus smaller) between $2$ numbers is $15$
If $n$ represents the larger number, which expression below represents the average (arithmetic mean) of the $2$ numbers?

$ F.\:\: 7.5 \\[1em] G.\:\: n + 7.5 \\[1em] H.\:\: n + 15 \\[1em] J.\:\: n - 15 \\[1em] K.\:\: n - 7.5 \\[1em] $

$ n = larger\:\:number...from\:\:the\:\:question \\[1em] Let\:\:p = smaller\:\:number \\[1em] n - p = 15...eqn.(1) \\[1em] Because\:\:the\:\:options\:\:are\:\:expressed\:\:in\:\:n,\:\:we\:\:have\:\:to\:\:solve\:\:for\:\:p \\[1em] n - 15 = p \\[1em] p = n - 15...eqn.(2) \\[1em] For\:\:n, p \\[1em] sample\:\:size = 2 \\[1em] sum = n + p \\[1em] \bar{x} = \dfrac{sum}{sample\:\:size} \\[2em] \bar{x} = \dfrac{n + p}{2} \\[2em] Substitute\:\:(n - 15)\:\:for\:\:p \\[1em] \bar{x} = \dfrac{n + (n - 15)}{2} \\[2em] \bar{x} = \dfrac{n + n - 15}{2} \\[2em] \bar{x} = \dfrac{2n - 15}{2} \\[2em] \bar{x} = \dfrac{2n}{2} - \dfrac{15}{2} \\[2em] \bar{x} = n - 7.5 $
(48.) JAMB $5, 8, 6$ and $k$ occur with frequencies $3, 2, 4$, and $1$ respectively and have a mean of $5.7$
Find the value of $k$

$ A.\:\: 4 \\[1em] B.\:\: 3 \\[1em] C.\:\: 2 \\[1em] D.\:\: 1 \\[1em] $

Numbers, $x$ Frequencies, $f$ $f * x$
$5$ $3$ $15$
$8$ $2$ $16$
$6$ $4$ $24$
$k$ $1$ $k$
$\Sigma f = 10$ $\Sigma fx = 55 + k$

$ \bar{x} = 5.7 \\[1em] \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[2em] 5.7 = \dfrac{55 + k}{10} \\[2em] 55 + k = 5.7 (10) \\[1em] 55 + k = 57 \\[1em] k = 57 - 55 \\[1em] k = 2 $
(49.) NSC A large company employs several people.
The table below shows the number of people employed in each position and the monthly salary paid to each person in that position.

POSITION NUMBER EMPLOYED IN POSITION MONTHLY SALARY PER PERSON (IN RAND)
Managing director $1$ $150\:000$
Director $2$ $100\:000$
Manager $2$ $75\:000$
Foreman $5$ $15\:000$
Skilled workers $30$ $10\:000$
Semi-skilled workers $40$ $7\:500$
Unskilled workers $65$ $6\:000$
Administration $5$ $5\:000$

(49.1) Calculate the total number of people employed at this company.

(49.2) Calculate the total amount needed to pay salaries for ONE month.

(49.3) Determine the mean monthly salary for an employee in this company.

(49.4) Is the mean monthly salary calculated in QUESTION (49.3) a good indicator of an employee's monthly salary? Motivate your answer.


$ (49.1) \\[1em] \Sigma People = 1 + 2 + 2 + 5 + 30 + 40 + 65 + 5 = 150 \\[1em] n = 150 \\[1em] (49.2) \\[1em] \Sigma Amount\:\:for\:\:one\:\:month \\[1em] = 1(150000) + 2(100000) + 2(75000) + 5(15000) + 30(10000) + 40(7500) + 65(6000) + 5(5000) \\[1em] = 150000 + 200000 + 150000 + 75000 + 300000 + 300000 + 390000 + 25000 \\[1em] = R1590000 \\[1em] \Sigma x = R1590000 \\[1em] (49.3) \\[1em] \bar{x} = \dfrac{\Sigma x}{n} \\[2em] \bar{x} = \dfrac{1590000}{150} \\[2em] \bar{x} = R10600 \\[1em] (49.4) \\[1em] R10600 \gt R10000 \\[1em] 30\:\:people\:\:earn\:\:R10000\:\:each \\[1em] 40\:\:people\:\:earn\:\:R7500\:\:each \\[1em] 65\:\:people\:\:earn\:\:R6000\:\:each \\[1em] 5\:\:people\:\:earn\:\:R5000\:\:each \\[1em] So \\[1em] 30 + 40 + 65 + 5 = 140\:\:people\:\:earn\:\:less\:\:than\:\:R10600 \\[1em] 140\:\:people\:\:out\:\:of\:\:150\:\:people\:\:earn\:\:below\:\:the\:\:average/mean\:\:amount\:\:of\:\:R10600 \\[1em] \dfrac{140}{150} * 100 = 93.33\% \\[2em] More\:\:than\:\:90\%\:\:earn\:\:below\:\:average \\[1em] Is\:\:that\:\:fair? \\[1em] The\:\:mean\:\:amount\:\:of\:\:R10600\:\:is\:\:NOT\:\:a\:\:good\:\:indicator\:\:of\:\:an\:\:employee's\:\:monthly\:\:salary \\[1em] OR \\[1em] R10600 \lt R15000 \\[1em] 5\:\:people\:\:earn\:\:R15000\:\:each \\[1em] 2\:\:people\:\:earn\:\:R75000\:\:each \\[1em] 2\:\:people\:\:earn\:\:R100000\:\:each \\[1em] 1\:\:person\:\:earn\:\:R150000 \\[1em] So \\[1em] 1 + 2 + 2 + 5 = 10\:\:people\:\:make\:\:more\:\:than\:\:R10600 \\[1em] 10\:\:people\:\:out\:\:of\:\:150\:\:people\:\:earn\:\:above\:\:the\:\:average/mean\:\:amount\:\:of\:\:R10600 \\[1em] \dfrac{10}{150} * 100 = 6.67\% \\[2em] Less\:\:than\:\:7\%\:\:earn\:\:above\:\:average \\[1em] Is\:\:that\:\:fair? \\[1em] The\:\:mean\:\:amount\:\:of\:\:R10600\:\:is\:\:NOT\:\:a\:\:good\:\:indicator\:\:of\:\:an\:\:employee's\:\:monthly\:\:salary \\[1em] $ Discuss the issue of $1%$ in some countries
Discuss the issue of corruption and poverty in almost all African countries
Discuss the evil of looting by politicians.
(50.) WASSCE-FM The table gives the distribution of heights, in meters, of $100$ students of the same age group.

Height $1.40 - 1.42$ $1.43 - 1.45$ $1.46 - 1.48$ $1.49 - 1.51$ $1.52 - 1.54$ $1.55 - 1.57$ $1.58 - 1.60$ $1.61 - 1.63$
Number of students $2$ $4$ $19$ $30$ $24$ $14$ $6$ $1$

(a) Calculate the mean height of the distribution.

(b) What is the probability that the height of a student selected at random is greater than the mean height of the distribution?


Check to make sure the gropued data has equal class intervals.
In other words, the class size must be the same for all the classes.

Height, $x$ Frequency, $f$ $x_{mid} = \dfrac{LCI + UCI}{2}$ $f * x_{mid}$
$1.40 - 1.42$ $2$ $\dfrac{1.4 + 1.42}{2} = \dfrac{2.82}{2} = 1.41$ $2.82$
$1.43 - 1.45$ $4$ $\dfrac{1.43 + 1.45}{2} = \dfrac{2.88}{2} = 1.44$ $5.76$
$1.46 - 1.48$ $19$ $\dfrac{1.46 + 1.48}{2} = \dfrac{2.94}{2} = 1.47$ $27.93$
$1.49 - 1.51$ $30$ $\dfrac{1.49 + 1.51}{2} = \dfrac{3}{2} = 1.5$ $45$
$1.52 - 1.54$ $24$ $\dfrac{1.52 + 1.54}{2} = \dfrac{3.06}{2} = 1.53$ $36.72$
$1.55 - 1.57$ $14$ $\dfrac{1.55 + 1.57}{2} = \dfrac{3.12}{2} = 1.56$ $21.84$
$1.58 - 1.6$ $6$ $\dfrac{1.58 + 1.6}{2} = \dfrac{3.18}{2} = 1.59$ $9.54$
$1.61 - 1.63$ $1$ $\dfrac{1.61 + 1.63}{2} = \dfrac{3.24}{2} = 1.62$ $1.62$
$\Sigma f = 100$ $\Sigma fx_{mid} = 151.23$

$ (a.) \\[1em] \bar{x} = \dfrac{\Sigma fx_{mid}}{\Sigma f} \\[2em] \bar{x} = \dfrac{151.23}{100} \\[2em] \bar{x} = 1.5123 \\[1em] Mean\:\:height = 1.5123\:meters \\[2em] (b) \\[1em] n(S) = \Sigma f = 100 \\[1em] Let\:\:the\:\:event\:\:of\:\:selecting\:\:a\:\:student\:\:whose\:\:height\:\:is\:\:greater\:\:than\:\:the\:\:mean\:\:height \\[1em] n(E) = 24 + 14 + 6 + 1 = 45 \\[1em] P(E) = \dfrac{n(E)}{n(S)} \\[2em] P(E) = \dfrac{45}{100} \\[2em] P(E) = \dfrac{9}{20} $
(51.) WASSCE-FM The table below shows the marks obtained by a group of students.

Marks $0 - 9$ $10 - 19$ $20 - 29$ $30 - 39$ $40 - 49$ $50 - 59$ $60 - 69$ $70 - 79$
Number of students $2$ $5$ $9$ $15$ $18$ $14$ $10$ $7$

(a) If a student is selected at random from the group, find the probability that this student scored at most $69$ marks.

(b) Calculate the median of the distribution.


The number of students is the frequency, $f$

$ (a.) \\[1em] \Sigma f = 2 + 5 + 9 + 15 + 18 + 14 + 10 + 7 \\[1em] \Sigma f = 80 \\[1em] n(S) = 80 \\[1em] At\:\:most\:\:69\:\:means\:\:\le 69 \\[1em] Let\:\:E = probability\:\:of\:\:selecting\:\:a\:\:student\:\:who\:\:scored\:\: \le 69 \\[1em] Number\:\:of\:\:students\:\:who\:\:scored\:\:less\:\:than\:\:or\:\:equal\:\:to\:\:69 \\[1em] = 2 + 5 + 9 + 15 + 18 + 14 + 10 \\[1em] = 73 \\[1em] n(E) = 73 \\[1em] P(E) = \dfrac{n(E)}{n(S)} \\[2em] P(E) = \dfrac{73}{80} \\[2em] $
Marks, $x$ Frequency, $F$ Class Boundaries Cumulative Frequency, $CF$
$0 - 9$ $2$ $0 - 9.5$ $2$
$10 - 19$ $5$ $9.5 - 19.5$ $2 + 5 = 7$
$20 - 29$ $9$ $19.5 - 29.5$ $7 + 9 = 16$
$30 - 39$ $15$ $29.5 - 39.5$ $16 + 15 = \color{darkblue}{31}$
$\color{darkblue}{40 - 49}$ $\color{darkblue}{18}$ $\color{darkblue}{39.5} - 49.5$ $31 + 18 = 49$
$50 - 59$ $14$ $49.5 - 59.5$ $49 + 14 = 63$
$60 - 69$ $10$ $59.5 - 69.5$ $63 + 10 = 73$
$70 - 79$ $7$ $69.5 - 79.5$ $73 + 7 = 80$
$\Sigma F = 80$

$ (b.) \\[1em] First:\:\:Determine\:\:the\:\:median\:\:class \\[1em] \dfrac{\Sigma F}{2} = \dfrac{80}{2} = 40 \\[2em] Begin\:\:from\:\:the\:\:first\:\:class \\[1em] Keep\:\:adding\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:40 \\[1em] 2 + 5 = 7 \\[1em] 7 + 9 = 16 \\[1em] 16 + 15 = 31 \\[1em] 31 + 18 = 49 \\[1em] 49 \:\:contains\:\: 40 \\[1em] STOP \\[1em] The\:\:median\:\:class = 40 - 49 \\[1em] \tilde{x} = LCB_{med} + \dfrac{CW}{f_{med}} * \left[\left(\dfrac{\Sigma F}{2}\right) - CF_{bmed}\right] \\[2em] med = 40 - 49 \\[1em] LCB_{med} = 39.5 \\[1em] CW = 49.5 - 39.5 = 10 \\[1em] f_{med} = 18 \\[1em] CF_{bmed} = 31 \\[1em] \implies \tilde{x} = 39.5 + \dfrac{10}{18} * \left[\left(\dfrac{80}{2}\right) - 31\right] \\[2em] \tilde{x} = 39.5 + \dfrac{10}{18} * (40 - 31) \\[2em] \tilde{x} = 39.5 + \dfrac{10}{18} * 9 \\[2em] \tilde{x} = 39.5 \dfrac{10}{2} \\[2em] \tilde{x} = 39.5 + 5 \\[1em] \tilde{x} = 44.5\:marks $
(52.) WASSCE-FM The table shows the distribution of the lengths of $20$ iron rods measured in metres.

Length (m) $1.0 - 1.1$ $1.2 - 1.3$ $1.4 - 1.5$ $1.6 - 1.7$ $1.8 - 1.9$
Frequency $2$ $3$ $8$ $5$ $2$

Using an assumed mean of $1.45$, calculate the mean of the distribution.


Check to make sure the gropued data has equal class intervals.
In other words, the class size must be the same for all the classes.
$AM = 1.45$

Length (m), $x$ Frequency, $f$ $x_{mid} = \dfrac{LCI + UCI}{2}$ $D = x_{mid} - AM$ $f * D$
$1.0 - 1.1$ $2$ $\dfrac{1.0 + 1.1}{2} = \dfrac{2.1}{2} = 1.05$ $1.05 - 1.45 = -0.4$ $2(-0.4) = -0.8$
$1.2 - 1.3$ $3$ $\dfrac{1.2 + 1.3}{2} = \dfrac{2.5}{2} = 1.25$ $1.25 - 1.45 = -0.2$ $3(-0.2) = -0.6$
$1.4 - 1.5$ $8$ $\dfrac{1.4 + 1.5}{2} = \dfrac{2.9}{2} = 1.45$ $1.45 - 1.45 = 0$ $8(0) = -0.8$
$1.6 - 1.7$ $5$ $\dfrac{1.6 + 1.7}{2} = \dfrac{3.3}{2} = 1.65$ $1.65 - 1.45 = 0.2$ $5(0.2) = 1.0$
$1.8 - 1.9$ $2$ $\dfrac{1.8 + 1.9}{2} = \dfrac{3.7}{2} = 1.85$ $1.85 - 1.45 = 0.4$ $2(0.4) = 0.8$
$\Sigma f = 20$ $\Sigma fD = 0.4$

$ \bar{x} = AM + \dfrac{\Sigma fD}{\Sigma f} \\[2em] \bar{x} = 1.45 + \dfrac{0.4}{20} \\[2em] \bar{x} = 1.45 + 0.02 \\[1em] Mean\:\:height = 1.47\:metres $