Solved Examples: Descriptive Statistics (All Measures)

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} $

$ 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 $

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.

(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 $

(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] $

(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.

$ Average\;\;number\;\;of\;\;moviegoers \\[3ex] = \dfrac{830 + 1650 + 2320 + 200}{4} \\[5ex] = \dfrac{5000}{4} \\[5ex] = 1,250 \\[3ex] $ The average number of moviegoers for each of the 4 categories is 1,250 moviegoers.

Let the adults that attended:
regular showings = x
matinees = y

A large theater complex surveyed 5,000 adults.

$ x + y = 5000 ...eqn.(1) \\[3ex] $ Tickets are $9.50 for all regular showings and $7.00 for matinees.
Suppose all the adults surveyed happened to attend 1 movie each in one particular week.
The total amount spent on tickets by those surveyed in that week was $44,000.00

$ 9.5x + 7y = 44000 \\[3ex] 95x + 70y = 440000 \\[3ex] 19x + 14y = 88000 ...eqn.(2) \\[3ex] \underline{Elimination\;\;Method} \\[3ex] 19 * eqn.(1) - eqn.(2) \implies \\[3ex] 19(x + y) - (19x + 14y) = 19(5000) - 88000 \\[3ex] 19x + 19y - 19x - 14y = 95000 - 88000 \\[3ex] 5y = 7000 \\[3ex] y = \dfrac{7000}{5} \\[5ex] y = 1400 \\[3ex] $ 1400 adults attended matinees that week.

We have about a minute to solve this question.
So, we shall find the sectorial angle for the first age group and then use the process of elimination to identify our answer
If necessary, we shall find the sectorial angle for the second age group and also eliminate options
We shall repeat this process until we get our answer

$ Sectorial\;\;\angle\;\;for\;\;each\;\;age\;\;group = \dfrac{frequency\;\;of\;\;the\;\;age\;\;group}{\Sigma f} * 100 \\[5ex] \underline{21-30\;\;Age\;\;Group} \\[3ex] Sectorial\;\;\angle = \dfrac{2750}{5000} * 100 \\[5ex] = 55\% \\[3ex] $ Eliminate Options B., D., and E.
Let us calculate the sectorial angle for the second age group

$ \underline{31-40\;\;Age\;\;Group} \\[3ex] Sectorial\;\;\angle = \dfrac{1225}{5000} * 100 \\[5ex] = 24.5\% \\[3ex] $ Eliminate Option C.
Option A. is the correct answer.

For those who just want to complete the rest of the age groups

$ \underline{41-50\;\;Age\;\;Group} \\[3ex] Sectorial\;\;\angle = \dfrac{625}{5000} * 100 \\[5ex] = 12.5\% \\[3ex] \underline{51\;\;or\;\;Older\;\;Age\;\;Group} \\[3ex] Sectorial\;\;\angle = \dfrac{400}{5000} * 100 \\[5ex] = 8\% \\[5ex] \underline{Check} \\[3ex] \Sigma Sectorial\;\; \angle = 55 + 24.5 + 12.5 + 8 = 100\% $

$ 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 $

$ \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 $

$ 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 $

Usually, the median and the mode are statistical measures that are not affected by outliers
Be it as it may, let us explain using the dataset that was given to us.
The dataset is already sorted
Hence, any data value (outlier) added to the dataset is before 60 or after 75 because outliers are extreme values
That outlier is not 60. It is not 75
But even if it is so (though it is not so); the median will not be affected
Presently, the median is 72 [(72 + 72) divided by 2...even sample size]
If an outlier is added, the median will still be 72...odd sample size
So, the median will not be affected
Presently, the mode is 73
If an outlier is added, the mode will still be 73
So, the median and the mode are not affected.

C. The center of the distribution of salaries for men in the country is greater than the center for women.
In a distribution of​ values, the typical value is given by the mean.
In this​ case, the average salary of a man is higher than that of a​ woman, so when comparing the distribution of​ men's salaries to​ women's salaries, the center of the distribution for men is greater than the center of the distribution for the women.

$ \underline{Domestic\;\;Gross\;\;Income\;(\$\;millions)} \\[3ex] (a.)\;\;sorted\;\;in\;\;ascending\;\;order \\[3ex] 384, 389, 408, 423, 434, 602 \\[5ex] (b.) \\[3ex] n = 6 \\[3ex] Median = \dfrac{408 + 423}{2} = \dfrac{831}{2} = 415.5\;million\;\$ \\[3ex] $ This means that about​ 50% of the top six based on Marvel comic book characters made more than four hundred and fifteen million and five hundred thouand dollars.

$ (c.) \\[3ex] n = 6 \\[3ex] 0.25 * 6 = 1.5 \approx 2 \\[3ex] Q_1 = 2nd\;\;value = 389\;million\;\$ \\[3ex] 0.75 * 6 = 4.5 \approx 5 \\[3ex] Q_3 = 5th\;\;value = 434\;million\;\$ \\[5ex] (d.) \\[3ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 434 - 389 = 45\;million\;\$ \\[3ex] $ Interpretation of IQR: The middle​ 50% of these 6 Marvel movies had domestic grosses that varied by as much as 45 million dollars.

$ (e.) \\[3ex] max = 602 \;million\;\$ \\[3ex] min = 384 \;million\;\$ \\[3ex] Range = max - min \\[3ex] Range = 602 - 384 = 218 \;million\;\$ \\[3ex] $ (f.) The interquartile range is preferred over the range as a measure of variability because it depends on many observations and is therefore more reliable.

Let us write Set A and Set B with numbers as elements of both sets (5 distinct numbers in each set), making sure that the least number in Set B is greater than the least number in Set A; while both sets have the same four numbers.
In other words, 4 of the 5 numbers in both sets are the same while the remaining number in Set B is greater than the remaining number in Set A.

Set A = {0, 2, 3, 4, 5}

Set B = {1, 2, 3, 4, 5}

$ \underline{Set\;A} \\[3ex] A = \{0, 2, 3, 4, 5\} \\[3ex] min = 0 \\[3ex] max = 5 \\[3ex] n = 5 \\[3ex] \Sigma x = 0 + 2 + 3 + 4 + 5 = 14 \\[3ex] mean = \dfrac{\Sigma x}{n} = \dfrac{14}{5} = 2.8 \\[5ex] median = 3 \\[3ex] range = max - min = 5 - 0 = 5 \\[5ex] \underline{Set\;B} \\[3ex] B = \{1, 2, 3, 4, 5\} \\[3ex] min = 1 \\[3ex] max = 5 \\[3ex] n = 5 \\[3ex] \Sigma x = 1 + 2 + 3 + 4 + 5 = 15 \\[3ex] mean = \dfrac{\Sigma x}{n} = \dfrac{15}{5} = 3 \\[5ex] median = 3 \\[3ex] range = max - min = 5 - 1 = 4 \\[3ex] $ Comparing Set A and Set B:
Mean of Set B is greater than mean of Set A
Median of Set B is the same as median of Set A
Range of Set B is less than range of Set A
Hence, based on the question we were asked, the only statistics of Set B that must be greater than Set A is the Mean.

A. The typical price for homes with pools is smaller than the typical price for homes without pools.

$ \underline{For\;\;16 - 25\;\;years} \\[3ex] Sectorial\;\;\angle\;\;for\;\;Company\;A \\[3ex] = \dfrac{Frequency\;\;for\;\;Company\;A}{\Sigma Frequency} * 360^\circ \\[5ex] = \dfrac{16}{80} * 360 \\[5ex] = 72^\circ $

$ 0\;hour \rightarrow 2\;students \\[3ex] 1\;hour \rightarrow 5\;students \\[3ex] 2\;hours \rightarrow 6\;students \\[3ex] 3\;hours \rightarrow 4\;students \\[3ex] 4\;hours \rightarrow 2\;students \\[3ex] 5\;hours \rightarrow 1\;student \\[3ex] Total\;\;number\;\;of\;\;students = \Sigma f = 2 + 5 + 6 + 4 + 2 + 1 = 20 \\[3ex] sectorial\;\angle \;\;for\;\;3\;\;hours \\[3ex] = \dfrac{number\;\;of\;\;students\;\;for\;\;3\;\;hours}{total\;\;number\;\;of\;\;students} * 360^\circ \\[5ex] = \dfrac{4}{20} * 360^\circ \\[5ex] = 72^\circ $

hours, $x$	number of students, $f$	$f * x$
$0$	$2$	$0$
$1$	$5$	$5$
$2$	$6$	$12$
$3$	$4$	$12$
$4$	$2$	$8$
$5$	$1$	$5$
	$\Sigma f = 20$	$\Sigma fx = 42$

$ \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] = \dfrac{42}{20} \\[5ex] = 2.1 $ The average number of hours for the 20 survey responses is 2.1 hours

Mean = $100$

Standard deviation = $15$

$1$ standard deviation = $15$

$1$ standard deviation above the mean = $15 + 100 = 115$

at least $1$ standard deviation above the mean is $\ge 115$

This means $115, 116, 117, 118, ...$

The lowest score a student could attain and still be accepted into the accounting class is $115$

The annual salaries of the 5 people at 12:00 pm is between $30,000 and $35,000
The mean is a measure of center: which is between $30,000 and $35,000
The median is also a measure of measure which is between $30,000 and $35,000
The range and the standard deviation will typically be less than $30,000
But then at 12:15 pm, an outlier of $1,000,000 is included in the data.
The mean will be highly affected...because it is not resistant to outliers.
The range and the standard deviation will also be highly affected.
The median will be the least affected because the 3rd and 4th data values (for a sample size of 6, the median is the average of the 3rd and 4th data values) are still within $30,000 and $35,000
In other words, the median is resistant to outliers.