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: Descriptive Statistics (All Measures)

Prerequisites:
Introductory Statistics
Solved Examples: Measures of Center
Solved Examples: Measures of Spread
Solved Examples: Measures of Location

Calculators:
Vertical Data Entry: Descriptive Statistics
Horizontal Data Entry: Descriptive Statistics

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.) WASSCE
Marks $1$ $2$ $3$ $4$ $5$
Number of students $m + 2$ $m - 1$ $2m - 3$ $m + 5$ $3m - 4$

The table shows the distribution of marks scored by some students in a test.
(1.) If the mean mark is $3\dfrac{6}{23}$, find the value of $m$

(2.) Find the:
(i) inter-quartile range;
(ii) probability of selecting a student who scored at least $4$ marks in the test.


Marks, $x$ Number of students, $f$ $f * x$
$1$ $m + 2$ $1(m + 2)$
$2$ $m - 1$ $2(m - 1)$
$3$ $2m - 3$ $3(2m - 3)$
$4$ $m + 5$ $4(m + 5)$
$5$ $3m - 4$ $5(3m - 4)$
$\Sigma f = 8m - 1$ $\Sigma fx = 28m - 9$

$ (1.) \\[1em] \bar{x} = 3\dfrac{6}{23} \\[2em] \Sigma f = (m + 2) + (m - 1) + (2m - 3) + (m + 5) + (3m - 4) \\[1em] = m + 2 + m - 1 + 2m - 3 + m + 5 + 3m - 4 \\[1em] = 8m - 1 \\[1em] \Sigma fx = 1(m + 2) + 2(m - 1) + 3(2m - 3) + 4(m + 5) + 5(3m - 4) \\[1em] = m + 2 + 2m - 2 + 6m - 9 + 4m + 20 + 15m - 20 \\[1em] = 28m - 9 \\[1em] \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] 3\dfrac{6}{23} = \dfrac{28m - 9}{8m - 1} \\[5ex] \dfrac{75}{23} = \dfrac{28m - 9}{8m - 1} \\[5ex] Cross\:\:Multiply \\[1em] 75(8m - 1) = 23(28m - 9) \\[1em] 600m - 75 = 644m - 207 \\[1em] -75 + 207 = 644m - 600m \\[1em] 132 = 44m \\[1em] 44m = 132 \\[1em] m = \dfrac{132}{44} \\[2em] m = 3 \\[1em] $ Let us update the table

Marks, $x$ Number of students, $f$
$1$ $m + 2$
$3 + 2$
$5$
means $1, 1, 1, 1, 1$
$2$ $m - 1$
$3 - 1$
$2$
means $2, 2$
$3$ $2m - 3$
$2(3) - 3$
$6 - 3$
$3$
means $3, 3, 3$
$4$ $m + 5$
$3 + 5$
$8$
means $4, 4, 4, 4, 4, 4, 4, 4$
$5$ $3m - 4$
$3(3) - 4$
$9 - 4$
$5$
means $5, 5, 5, 5, 5$
$\Sigma f$
$8m - 1$
$8(3) - 1$
$24 - 1$
$23$
means $\Sigma f = n = 23$

$ (2.) \\[1em] The\:\:dataset\:\:is \\[1em] 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5 \\[1em] (i) \\[1em] IQR = Q_3 - Q_1 \\[1em] Q_3 = 75th\:\:percentile\:\:of\:\:n \\[1em] = \dfrac{75}{100} * 23 \\[2em] = 0.75 * 23 \\[1em] = 17.25th\:position...not\:\:an\:\:integer...round\:\:up \\[1em] = 18th\:position \\[1em] = 4 \\[1em] Q_1 = 25th\:\:percentile\:\:of\:\:n \\[1em] = \dfrac{25}{100} * 23 \\[2em] = 0.25 * 23 \\[1em] = 5.75th\:position...not\:\:an\:\:integer...round\:\:up \\[1em] = 6th\:position \\[1em] = 2 \\[1em] \implies IQR = 4 - 2 \\[1em] IQR = 2 \\[1em] (ii) \\[1em] At\:\:least\:\:4\:\:means \ge 4 \\[1em] Let\:\:the\:\:event\:\:of\:\:selecting\:\:a\:\:student\:\:who\:\:at\:\:least\:\:4 = E \\[1em] n(E) = 8 + 5 = 13 \\[1em] n(S) = 23 \\[1em] P(E) = \dfrac{n(E)}{n(S)} \\[2em] P(E) = \dfrac{13}{23} $
(2.) JAMB The mean and the range of the set of numbers $0.20, 1.00, 0.90, 1.40, 0.80, 0.80, 1.20, 1.10$ are $m$ and $r$ respectively.
Find $m + r$

$ A.\;\; 1.11 \\[3ex] B.\;\; 1.65 \\[3ex] C.\;\; 1.85 \\[3ex] D.\;\; 2.45 \\[3ex] $

$ n = sample\;\;size = 8 \\[3ex] \Sigma x = 1.20 + 1.00 + 0.90 + 1.40 + 0.80 + 0.80 + 1.20 + 1.10 \\[3ex] = 8.4 \\[3ex] \underline{Mean} \\[3ex] m = \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] = \dfrac{8.4}{8} \\[5ex] = 1.05 \\[3ex] \underline{Range} \\[3ex] minimum = 0.20 \\[3ex] maximum = 1.40 \\[3ex] r = Range = maximum - minimum \\[3ex] = 1.40 - 0.80 \\[3ex] = 0.6 \\[3ex] m + r \\[3ex] = 1.05 + 0.6 \\[3ex] = 1.65 $
(3.) NSC In the grid below $a, b, c, d, e, f$ and $g$ represent values in a data set written in an increasing order.
No value in the data set is repeated.

$a$ $b$ $c$ $d$ $e$ $f$ $g$

Determine the value of $a, b, c, d, e, f$ and $g$ if:
  • The maximum value is $42$
  • The range is $35$
  • The median is $23$
  • *The difference between the median and the upper quartile is $14$ (improper way as it was written in the past question)*
  • The difference between the upper quartile and the median is $14$ (proper way to write it, so please write it as this)
  • The interquartile range is $22$
  • $e = 2c$
  • The mean is $25$

Increasing order means Ascending order
It means that the values are ordered from least to greatest
This means that $a$ is the least and $g$ is the greatest

$ \underline{First\:\:Point} \\[1em] max = g = 42 \\[1em] \underline{Second\:\:Point} \\[1em] range = max - min \\[1em] 35 = 42 - min \\[1em] min = 42 - 35 \\[1em] min = 7 \\[1em] a = min \\[1em] a = 7 \\[1em] \underline{Third\:\:Point} \\[1em] n = 7 \\[1em] median = middle = Q_2 = 4th\:\:position = d \\[1em] 23 = Q_2 = d \\[1em] d = Q_2 = 23 \\[1em] \underline{Fourth\:\:Point} \\[1em] Q_3 - Q_2 = 14 \\[1em] Q_3 = 75th\:\:percentile\:\:of\:\:n \\[1em] = \dfrac{75}{100} * 7 \\[2em] = 0.75 * 7 \\[1em] = 5.25th\:position...not\:\:an\:\:integer...round\:\:up \\[1em] = 6th\:position \\[1em] = f \\[1em] Q_3 = f \\[1em] \implies f - 23 = 14 \\[1em] f = 14 + 23 \\[1em] f = 37 \\[1em] \underline{Fifth\:\:Point} \\[1em] IQR = Q_3 - Q_1 \\[1em] 22 = 37 - Q_1 \\[1em] Q_1 = 37 - 22 \\[1em] Q_1 = 15 \\[1em] Also \\[1em] Q_1 = 25th\:\:percentile\:\:of\:\:n \\[1em] = \dfrac{25}{100} * 7 \\[2em] = 0.25 * 7 \\[1em] = 1.75th\:position...not\:\:an\:\:integer...round\:\:up \\[1em] = 2nd\:position \\[1em] = b \\[1em] Q_1 = b = 15 \\[1em] b = 15 \\[1em] \underline{Sixth\:\:Point} \\[1em] e = 2c...eqn.(1) \\[1em] \underline{Seventh\:\:Point} \\[1em] \bar{x} = 25 \\[1em] \bar{x} = \dfrac{\Sigma x}{n} \\[2em] \Sigma x = a + b + c + d + e + f + g \\[1em] \Sigma x = 7 + 15 + c + 23 + e + 37 + 42 \\[1em] \Sigma x = 124 + c + e \\[1em] Substitute\:\:e = 2c...from\:\:eqn.(1) \\[1em] \Sigma x = 124 + c + 2c \\[1em] \Sigma x = 124 + 3c \\[1em] \implies 25 = \dfrac{124 + 3c}{7} \\[2em] 25(7) = 124 + 3c \\[1em] 175 = 124 + 3c \\[1em] 124 + 3c = 175 \\[1em] 3c = 175 - 124 \\[1em] 3c = 51 \\[1em] c = \dfrac{51}{3} \\[2em] c = 17 \\[1em] e = 2c = 2(17) \\[1em] e = 34 \\[1em] $ The completed grid is:
$7$ $15$ $17$ $23$ $34$ $37$ $42$

The numbers are in ascending order.
None of the values are repeated.
(4.) WASSCE The table below shows the number of children per family in a community.

Number of children $0$ $1$ $2$ $3$ $4$ $5$
Number of families $3$ $5$ $7$ $4$ $3$ $2$


(a.) Find the:
(i) mode
(ii) third quartile
(iii) probability that a family has at least $2$ children

(b.) If a pie chart were to be drawn for the data, what would be the sectorial angle representing families with one child?


(a.)
Number of children are the data values, $x$
Number of families are the frequencies, $f$
(i)
The mode is the number of children in the highest number of families
Highest number of families is $7$
The number of children in $7$ families is $2$
$Mode = 2$

$ (ii) \\[1em] n = \Sigma f = 3 + 5 + 7 + 4 + 3 + 2 = 24 \\[1em] Q_3 = 75th\:\:percentile\:\:of\:\:n \\[1em] = \dfrac{75}{100} * 24 \\[2em] = 0.75 * 24 \\[1em] = 18th\:position...\:\:an\:\:integer...so \\[1em] find\:\:the\:\:18th\:\:position...and \\[1em] round\:\:up...find\:\:the\:\:19th\:\:position...and \\[1em] find\:\:the\:\:average \\[1em] Two\:\:methods \\[1em] \underline{First\:\:Method: Recommended} \\[1em] 18th\:\:position \\[1em] Add\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:18 \\[1em] 3 + 5 = 8 \\[1em] 8 + 7 = 15 \\[1em] 15 + 4 = 19...STOP \\[1em] Which\:\:children? \\[1em] 18th\:position = 3 \\[1em] Also:\:\: 118th\:position = 3 \\[1em] \therefore Q_3 = \dfrac{18th\:position + 19th\:position}{2} \\[2em] Q_3 = \dfrac{3 + 3}{2} \\[2em] Q_3 = \dfrac{6}{2} \\[2em] Q_3 = 3 \\[1em] OR \\[1em] \underline{Second\:\:Method: Still\:\:Okay} \\[1em] List\:\:the\:\:entire\:\:data\:\:values \\[1em] 0\:\:occured\:\:3\:\:times \implies 0, 0, 0 \\[1em] 1\:\:occured\:\:5\:\:times \implies 1, 1, 1, 1, 1 \\[1em] 2\:\:occured\:\:7\:\:times \implies 2, 2, 2, 2, 2, 2, 2 \\[1em] 3\:\:occured\:\:4\:\:times \implies 3, 3, 3, 3 \\[1em] 4\:\:occured\:\:3\:\:times \implies 4, 4, 4 \\[1em] 5\:\:occured\:\:2\:\:times \implies 5, 5 \\[1em] The\:\:raw\:\:dataset\:\:is \\[1em] 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5 \\[1em] 18th\:position = 3 \\[1em] 18th\:position = 3 \\[1em] Q_3 = \dfrac{18th\:position + 19th\:position}{2} \\[2em] Q_3 = \dfrac{3 + 3}{2} \\[2em] Q_3 = \dfrac{6}{2} \\[2em] Q_3 = 3 \\[1em] (iii) \\[1em] At\:\:least\:\:2 \implies \gt 2 \\[1em] Let\:\:E\:\:be\:\:the\:\:event\:\:that\:\:a\:\:family\:\:has\:\:2\:\:children \\[1em] 7\:\:families\:\:have\:\:2\:\:children \\[1em] 4\:\:families\:\:have\:\:3\:\:children \\[1em] 3\:\:families\:\:have\:\:4\:\:children \\[1em] 2\:\:families\:\:have\:\:5\:\:children \\[1em] n(E) = 7 + 4 + 3 + 2 = 16 \\[1em] n(S) = \Sigma f = 24 \\[1em] P(E) = \dfrac{n(E)}{n(S)} \\[2em] P(E) = \dfrac{16}{24} \\[2em] P(E) = \dfrac{2}{3} \\[2em] (b.) \\[1em] Number\:\:of\:\:families\:\:with\:\:one\:\:child = 5 \\[1em] Total\:\:number\:\:of\:\:families = n = \Sigma f = 24 \\[1em] Total\:\:angle\:\:in\:\:a\:\:circle = 360^\circ \\[1em] Sectoral\: \angle\:\: for\:\:families\:\:with\:\:one\:\:child \\[1em] = \dfrac{5}{24} * 360 \\[2em] = 5 * 15 \\[1em] = 75^\circ $
(5.) WASSCE
Class Interval Frequency
$60-64$ $2$
$65-69$ $3$
$70-74$ $6$
$75-79$ $11$
$80-84$ $8$
$85-89$ $7$
$90-94$ $2$
$95-99$ $1$

The table shows the distribution of marks scored by students in an examination.
Calculate, correct to 2 decimal places, the
(a.) mean
(b.) standard deviation of the distribution.


(a.)

$ Mean = \dfrac{\Sigma fx_{mid}}{\Sigma f} \\[5ex] $
Class Interval, $x$ Class Midpoint, $x_{mid}$ Frequency, $f$ $f * x_{mid}$
$60-64$ $\dfrac{60 + 64}{2} = \dfrac{124}{2} = 62$ $2$ $124$
$65-69$ $\dfrac{65 + 69}{2} = \dfrac{134}{2} = 67$ $3$ $201$
$70-74$ $\dfrac{70 + 74}{2} = \dfrac{144}{2} = 72$ $6$ $432$
$75-79$ $\dfrac{75 + 79}{2} = \dfrac{154}{2} = 77$ $11$ $847$
$80-84$ $\dfrac{80 + 84}{2} = \dfrac{164}{2} = 82$ $8$ $656$
$85-89$ $\dfrac{85 + 89}{2} = \dfrac{174}{2} = 87$ $7$ $609$
$90-94$ $\dfrac{90 + 94}{2} = \dfrac{184}{2} = 92$ $2$ $184$
$95-99$ $\dfrac{95 + 99}{2} = \dfrac{194}{2} = 97$ $1$ $97$
$\Sigma f = 40$ $\Sigma fx_{mid} = 3150$

$ \bar{x} \\[3ex] = \dfrac{\Sigma fx_{mid}}{\Sigma f} \\[5ex] = \dfrac{3150}{40} \\[5ex] = 78.75 \\[3ex] $ (b.)
There are two methods/formulas to solve this question.
Use any method you prefer.

First Method: Mean is used

$ Sample\;\;Standard\;\;Deviation = \sqrt{\dfrac{\Sigma f(x_{mid} - \bar{x})^2}{\Sigma f - 1}} \\[5ex] Population\;\;Standard\;\;Deviation = \sqrt{\dfrac{\Sigma f(x_{mid} - \bar{x})^2}{\Sigma f}} \\[5ex] $
$x_{mid} - \bar{x}$ $(x_{mid} - \bar{x})^2$ $f * (x_{mid} - \bar{x})^2$
$62 - 78.75 = -16.75$ $(-16.75)^2 = 280.5625$ $2 * 280.5625 = 561.125$
$67 - 78.75 = -11.75$ $(-11.75)^2 = 138.0625$ $3 * 138.0625 = 414.1875$
$72 - 78.75 = -6.75$ $(-6.75)^2 = 45.5625$ $6 * 45.5625 = 273.375$
$77 - 78.75 = -1.75$ $(-1.75)^2 = 3.0625$ $11 * 3.0625 = 33.6875$
$82 - 78.75 = 3.25$ $(3.25)^2 = 10.5625$ $8 * 10.5625 = 84.5$
$87 - 78.75 = 8.25$ $(8.25)^2 = 68.0625$ $7 * 68.0625 = 476.4375$
$92 - 78.75 = 13.25$ $(13.25)^2 = 175.5625$ $2 * 175.5625 = 351.125$
$97 - 78.75 = 18.25$ $(18.25)^2 = 333.0625$ $1 * 333.0625 = 333.0625$
$\Sigma f * (x_{mid} - \bar{x})^2 = 2527.5$

$ Sample\;\;Standard\;\;Deviation \\[3ex] = \sqrt{\dfrac{\Sigma f(x_{mid} - \bar{x})^2}{\Sigma f - 1}} \\[5ex] = \sqrt{\dfrac{2527.5}{40 - 1}} \\[5ex] = \sqrt{\dfrac{2527.5}{39}} \\[5ex] = \sqrt{64.80769231} \\[3ex] = 8.050322497 \\[3ex] \approx 8.05 \\[5ex] Population\;\;Standard\;\;Deviation \\[3ex] = \sqrt{\dfrac{\Sigma f(x_{mid} - \bar{x})^2}{\Sigma f}} \\[5ex] = \sqrt{\dfrac{2527.5}{40}} \\[5ex] = \sqrt{63.1875} \\[3ex] = 7.949056548 \\[3ex] \approx 7.95 \\[3ex] $ Second Method: Mean is not used

$ Sample\;\;Standard\;\;Deviation = \sqrt{\dfrac{\Sigma f(\Sigma fx_{mid}^2) - (\Sigma fx_{mid})^2}{\Sigma f(\Sigma f - 1)}} \\[5ex] Population\;\;Standard\;\;Deviation = \sqrt{\dfrac{\Sigma f(\Sigma fx_{mid}^2) - (\Sigma fx_{mid})^2}{\Sigma f(\Sigma f)}} \\[5ex] $
$x_{mid}$ $f$ $f * x_{mid}$ $(x_{mid})^2$ $f * (x_{mid})^2$
$62$ $2$ $124$ $3844$ $7688$
$67$ $3$ $201$ $4489$ $13467$
$72$ $6$ $432$ $5184$ $31104$
$77$ $11$ $847$ $5929$ $65219$
$82$ $8$ $656$ $6724$ $53792$
$87$ $7$ $609$ $7569$ $52983$
$92$ $2$ $184$ $8464$ $16928$
$97$ $1$ $97$ $9409$ $9409$
$\Sigma f * x_{mid} = 3150$ $\Sigma f * (x_{mid})^2 = 250590$

$ Sample\;\;Standard\;\;Deviation \\[3ex] = \sqrt{\dfrac{\Sigma f(\Sigma fx_{mid}^2) - (\Sigma fx_{mid})^2}{\Sigma f(\Sigma f - 1)}} \\[5ex] = \sqrt{\dfrac{40(250590) - (3150)^2}{40(40 - 1)}} \\[5ex] = \sqrt{\dfrac{10023600 - 9922500}{40(39)}} \\[5ex] = \sqrt{\dfrac{101100}{1560}} \\[5ex] = \sqrt{64.80769231} \\[3ex] = 8.050322497 \\[3ex] \approx 8.05 \\[5ex] Population\;\;Standard\;\;Deviation \\[3ex] = \sqrt{\dfrac{\Sigma f(\Sigma fx_{mid}^2) - (\Sigma fx_{mid})^2}{\Sigma f(\Sigma f)}} \\[5ex] = \sqrt{\dfrac{40(250590) - (3150)^2}{40(40)}} \\[5ex] = \sqrt{\dfrac{10023600 - 9922500}{1600}} \\[5ex] = \sqrt{\dfrac{101100}{1600}} \\[5ex] = \sqrt{63.1875} \\[3ex] = 7.949056548 \\[3ex] \approx 7.95 \\[3ex] $
(6.) curriculum.gov.mt These are the marks that Stephanie scored in her Maths tests:
68,     58,     61,     75,     63.
These are the marks that Stephanie scored in her Science tests:
81,     77,     68,     76.

(a.) Complete the following table:
Mean Median Range
Maths
Science


(b.) In which subject did Stephanie do better, Maths or Science?
Give a reason for your answer.


(a.)
Mean Median Range
Maths $ Mean \\[3ex] = \dfrac{68 + 58 + 61 + 75 + 63}{5} \\[5ex] = \dfrac{325}{5} \\[5ex] = 65 $ Sorted in Ascending Order:
58, 61, 63, 68, 75
Median = 63
Maximum = 75
Minimum = 58
Range = 75 - 58
Range = 17
Science $ Mean \\[3ex] = \dfrac{81 + 77 + 68 + 76}{4} \\[5ex] = \dfrac{302}{4} \\[5ex] = 75.5 $ Sorted in Ascending Order:
68, 76, 77, 81

$ Median \\[3ex] = \dfrac{76 + 77}{2} \\[5ex] = \dfrac{153}{2} \\[5ex] = 76.5 $
Maximum = 81
Minimum = 68
Range = 81 - 68
Range = 13

(b.) Stephanie did better in Science because her mean and median scores in Science are higher than her mean and median scores in Maths.
(7.) WASSCE-FM The table shows the distribution of marks obtained by students in an examination.

Marks $50 - 54$ $55 - 59$ $60 - 64$ $65 - 69$ $70 - 74$ $75 - 79$ $80 - 84$ $85 - 89$
Frequency $5$ $15$ $20$ $28$ $12$ $9$ $7$ $4$

Using an assumed mean (AM) of $67$, calculate, correct to one decimal place, the:
(a.) mean;
(b.) standard deviation;
of the distribution.


Check to make sure the grouped data has equal class intervals.
In other words, the class size must be the same for all the classes.
$AM = 67$

(a.)
Marks, $x$ Frequency, $f$ $x_{mid} = \dfrac{LCI + UCI}{2}$ $D = x_{mid} - AM$ $f * D$
$50 - 54$ $5$ $\dfrac{50 + 54}{2} = \dfrac{104}{2} = 52$ $52 - 67 = -15$ $5(-15) = -75$
$55 - 59$ $15$ $\dfrac{55 + 59}{2} = \dfrac{114}{2} = 57$ $57 - 67 = -10$ $15(-10) = -150$
$60 - 64$ $20$ $\dfrac{60 + 64}{2} = \dfrac{124}{2} = 62$ $62 - 67 = -5$ $20(-5) = -100$
$65 - 69$ $28$ $\dfrac{65 + 69}{2} = \dfrac{134}{2} = 67$ $67 - 67 = 0$ $28(0) = 0$
$70 - 74$ $12$ $\dfrac{70 + 74}{2} = \dfrac{144}{2} = 72$ $72 - 67 = 5$ $12(5) = 60$
$75 - 79$ $9$ $\dfrac{75 + 79}{2} = \dfrac{154}{2} = 77$ $77 - 67 = 10$ $9(10) = 90$
$80 - 84$ $7$ $\dfrac{80 + 84}{2} = \dfrac{164}{2} = 82$ $82 - 67 = 15$ $7(15) = 105$
$85 - 89$ $4$ $\dfrac{85 + 89}{2} = \dfrac{174}{2} = 87$ $87 - 67 = 20$ $4(20) = 80$
$\Sigma f = 100$ $\Sigma fD = 10$

$ \bar{x} = AM + \dfrac{\Sigma fD}{\Sigma f} \\[5ex] \bar{x} = 67 + \dfrac{10}{100} \\[5ex] \bar{x} = 67 + 0.1 \\[3ex] \bar{x} = 67.1 \\[5ex] $ (b.)
Using Assumed Mean:

$ Sample\;\;Standard\;\;Deviation = \sqrt{\dfrac{\Sigma fD^2}{\Sigma f - 1} - \left(\dfrac{\Sigma fD}{\Sigma f - 1}\right)^2} \\[5ex] Population\;\;Standard\;\;Deviation = \sqrt{\dfrac{\Sigma fD^2}{\Sigma f} - \left(\dfrac{\Sigma fD}{\Sigma f}\right)^2} \\[5ex] $
$f$ $D$ $f * D$ $D^2$ $f * D^2$
$5$ $-15$ $-75$ $225$ $1125$
$15$ $-10$ $-150$ $100$ $1500$
$20$ $-5$ $-100$ $25$ $500$
$28$ $0$ $0$ $0$ $0$
$12$ $5$ $60$ $25$ $300$
$9$ $10$ $90$ $100$ $900$
$7$ $15$ $105$ $225$ $1575$
$4$ $20$ $80$ $400$ $1600$
$\Sigma f = 100$ $\Sigma fD = 10$ $\Sigma fD^2 = 7500$


$ Sample\;\;Standard\;\;Deviation \\[3ex] = \sqrt{\dfrac{\Sigma fD^2}{\Sigma f - 1} - \left(\dfrac{\Sigma fD}{\Sigma f - 1}\right)^2} \\[5ex] = \sqrt{\dfrac{7500}{100 - 1} - \left(\dfrac{10}{100 - 1}\right)^2} \\[5ex] = \sqrt{\dfrac{7500}{99} - \left(\dfrac{10}{99}\right)^2} \\[5ex] = \sqrt{\dfrac{7500}{99} - \dfrac{100}{99^2}} \\[5ex] = \sqrt{\dfrac{99(7500) - 100}{99^2}} \\[5ex] = \sqrt{\dfrac{742500 - 100}{9801}} \\[5ex] = \sqrt{\dfrac{742400}{9801}} \\[5ex] = \sqrt{75.74737272} \\[3ex] = 8.703296658 \\[3ex] \approx 8.7 \\[5ex] Population\;\;Standard\;\;Deviation \\[3ex] = \sqrt{\dfrac{\Sigma fD^2}{\Sigma f} - \left(\dfrac{\Sigma fD}{\Sigma f}\right)^2} \\[5ex] = \sqrt{\dfrac{7500}{100} - \left(\dfrac{10}{100}\right)^2} \\[5ex] = \sqrt{75 - (0.1)^2} \\[3ex] = \sqrt{75 - 0.01} \\[3ex] = \sqrt{74.99} \\[3ex] = 8.659676668 \\[3ex] \approx 8.7 $
(9.)

(10.) GCSE A set of data has
mean = 30
median = 25
standard deviation = 4
Circle the value of the skew for the data.

$ Use\;\;skew = \dfrac{3(mean - median)}{standard\;\;deviation} \\[5ex] $ -11.25     1.25     3.75     16.25


$ mean = 30 \\[3ex] median = 25 \\[3ex] standard\;\;deviation = 4 \\[3ex] skew \\[3ex] = \dfrac{3(mean - median)}{standard\;\;deviation} \\[5ex] = \dfrac{3(30 - 25)}{4} \\[5ex] = \dfrac{3(5)}{4} \\[5ex] = \dfrac{15}{4} \\[5ex] = 3.75 $
(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.) WASSCE The range of the numbers $p, 6, 8, 11\:and\:q$ arranged in ascending order is $16$.
If the mean is $9$, find the value of $(2p + q)$


$ p, 6, 8, 11, q ...ascending\:order \\[3ex] min = p \\[3ex] max = q \\[3ex] R = max - min = 16 \\[3ex] \rightarrow q - p = 16 \\[3ex] -p + q = 16 ...eqn.(1) \\[3ex] \bar{x} = \dfrac{p + 6 + 8 + 11 + q}{5} = 9 \\[5ex] \dfrac{p + q + 25}{5} = 9 \\[5ex] p + q + 25 = 9(5) \\[3ex] p + q + 25 = 45 \\[3ex] p + q = 45 - 25 \\[3ex] p + q = 20 ...eqn.(2) \\[3ex] eqn.(1) + eqn.(2) \rightarrow q + q = 16 + 20 \\[3ex] 2q = 36 \\[3ex] q = \dfrac{36}{2} \\[5ex] q = 18 \\[3ex] eqn.(1) - eqn.(2) \rightarrow -p - p = 16 - 20 \\[3ex] -2p = -4 \\[3ex] p = \dfrac{-4}{-2} \\[5ex] p = 2 \\[3ex] \underline{Check} \\[3ex] q - p = 18 - 2 = 16 \checkmark \\[3ex] \dfrac{p + 6 + 8 + 11 + q}{5} = \dfrac{2 + 6 + 8 + 11 + 18}{5} = \dfrac{45}{5} = 9 \checkmark \\[5ex] \therefore 2p + q = 2(2) + 18 = 4 + 18 = 22 $
(13.)

(14.)


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.