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Statistics and Probability

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These are the solutions to the SACE past questions on Statistics and Probability.
The TI-84 Plus CE shall be used for applicable questions.
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Grouped Data

$ \underline{\text{Class Size or Class Width}} \\[3ex] (1.)\;\; Class\:\:Width = \dfrac{Maximum - Minimum}{Number\:\:of\:\:classes} \\[5ex] (2.)\;\; Class\:\:Width = LCI\:\:of\:\:2nd\:\:Class - LCI\:\:of\:\:1st\:\:Class \\[3ex] (3.)\;\; Class\:\:Width = UCI\:\:of\:\:2nd\:\:Class - UCI\:\:of\:\:1st\:\:Class \\[3ex] (4.)\;\; Class\:\:Width = UCB\:\:of\:\:a\:\:class - LCB\:\:of\:\:the\:\:same\:\:class \\[3ex] (5.)\;\; Class\:\:Width = LCB\:\:of\:\:a\:\:Class - LCB\:\:of\:\:previous\:\:class \\[5ex] \underline{\text{Frequency Density}} \\[3ex] (6.)\;\; \text{Frequency Density} = \dfrac{\text{Frequency}}{\text{Class Width}} \\[7ex] \underline{\text{Class Midpoints or Class Marks}} \\[3ex] (7.)\;\; Class\:\:Width = LCB\:\:of\:\:a\:\:Class - LCB\:\:of\:\:previous\:\:class \\[5ex] \underline{\text{Class Boundaries}} \\[3ex] (8.)\;\; Lower\:\:Class\:\:Boundary\:\:of\:\:a\:\:class = \dfrac{LCI\:\:of\:\:that\:\:class + UCI\:\:of\:\:previous/preceding\:\:class}{2} \\[5ex] (9.)\;\; Upper\:\:Class\:\:Boundary\:\:of\:\:a\:\:class = \dfrac{UCI\:\:of\:\:that\:\:class + LCI\:\:of\:\:next/succeeding\:\:class}{2} \\[5ex] $ (10.) Shortcut for Class Boundaries
If the class intervals are integers:
Lower Class Boundary = Lower Class Interval − 0.5
Upper Class Boundary = Upper Class Interval + 0.5

If the class intervals are decimals in one decimal place:
Lower Class Boundary = Lower Class Interval − 0.05
Upper Class Boundary = Upper Class Interval + 0.05

If the class intervals are decimals in two decimal places:
Lower Class Boundary = Lower Class Interval − 0.005
Upper Class Boundary = Upper Class Interval + 0.005

...and so on and so forth.

$ \underline{\text{Relative Frequency}} \\[3ex] (11.)\;\; RF\:\:of\:\:a\:\:class = \dfrac{Frequency\:\:of\:\:that\:\:class}{\Sigma Frequency} \\[7ex] \underline{\text{Cumulative Frequency}} \\[3ex] (12.)\;\; CF\:\:of\:\:1st\:\:Class = Frequency\:\:of\:\:1st\:\:Class \\[3ex] CF\:\:of\:\:2nd\:\:Class = Frequency\:\:of\:\:1st\:\:Class + Frequency\:\:of\:\:2nd\:\:Class \\[3ex] CF\:\:of\:\:3rd\:\:Class = Frequency\:\:of\:\:1st\:\:Class + Frequency\:\:of\:\:2nd\:\:Class + Frequency\:\:of\:\:3rd\:\:Class \\[3ex] CF = CF\:\:of\:\:Last\:\:Class = \Sigma Frequency $


Measures of Center: Raw Data and Ungrouped Data

$ \underline{Sample\:\:Mean} \\[3ex] (1.)\:\: \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] (2.)\:\: n = \Sigma f \\[3ex] (3.)\:\: \bar{x} = \dfrac{\Sigma fx}{\Sigma f} \\[5ex] \underline{Given\:\:an\:\:Assumed\:\:Mean} \\[3ex] (4.)\:\: D = x - AM \\[3ex] (5.)\:\: \bar{x} = AM + \dfrac{\Sigma D}{n} \\[5ex] (6.)\:\: \bar{x} = AM + \dfrac{\Sigma fD}{\Sigma f} \\[7ex] \underline{Population\:\:Mean} \\[3ex] (7.)\:\: \mu = \dfrac{\Sigma x}{N} \\[5ex] (8.)\:\: N = \Sigma f \\[3ex] \underline{Given\:\:an\:\:Assumed\:\:Mean} \\[3ex] (9.)\:\: D = x - AM \\[3ex] (10.)\:\: \mu = AM + \dfrac{\Sigma D}{N} \\[5ex] (11.)\:\: \mu = AM + \dfrac{\Sigma fD}{\Sigma f} \\[7ex] \underline{Median} \\[3ex] (12.)\:\: \tilde{x} = \left(\dfrac{\Sigma f + 1}{2}\right)th \:\:for\:\:sorted\:\:odd\:\:sample\:\:size \\[5ex] (13.)\:\: \tilde{x} = \left(\dfrac{\Sigma f}{2}\right)th \:\:for\:\:sorted\:\:even\:\:sample\:\:size \\[7ex] \underline{Mode} \\[3ex] (14.)\:\: Mode = x-value(s) \:\;with\:\:highest\:\:frequency \\[5ex] \underline{Midrange} \\[3ex] (15.)\:\: x_{MR} = \dfrac{min + max}{2} \\[5ex] \underline{Geometric\;\;Mean} \\[3ex] (16.)\;\; GM = \sqrt[n]{\prod\limits_{x=1}^n x} $


Measures of Center: Grouped Data

$ \underline{Class\:\:Midpoint} \\[3ex] (1.)\:\: x_{mid} = \dfrac{LCL + UCL}{2} \\[7ex] Equal\:\:Class\:\:Intervals\:(Same\:\:Class\:\:Size) \\[3ex] \underline{Mean} \\[3ex] (2.)\:\: \bar{x} = \dfrac{\Sigma fx_{mid}}{\Sigma f} \\[7ex] Equal\:\:Class\:\:Intervals\:(Same\:\:Class\:\:Size) \\[3ex] \underline{Given\:\:an\:\:Assumed\:\:Mean} \\[3ex] (3.)\:\: D = x_{mid} - AM \\[3ex] (4.)\:\: \bar{x} = AM + \dfrac{\Sigma fD}{\Sigma f} \\[7ex] \underline{Median} \\[3ex] (5.)\:\: \tilde{x} = LCB_{med} + \dfrac{CW}{f_{med}} * \left[\left(\dfrac{\Sigma f}{2}\right) - CF_{bmed}\right] \\[7ex] \underline{Mode} \\[3ex] (6.)\:\: \widehat{x} = LCB_{mod} + CW * \left[\dfrac{f_{mod} - f_{bmod}}{(f_{mod} - f_{bmod}) + (f_{mod} - f_{amod})}\right] $


Measures of Spread: Raw Data and Ungrouped Data

$ \underline{Range} \\[3ex] (1.)\:\: Range = max - min \\[3ex] \underline{Using\;\;Assumed\;\;Mean} \\[3ex] (2.)\;\; D = x - AM \\[5ex] \underline{Sample\:\:Variance} \\[3ex] \color{red}{First\:\:Formula} \\[3ex] (3.)\:\: s^2 = \dfrac{\Sigma(x - \bar{x})^2}{n - 1} \\[5ex] (4.)\:\: s^2 = \dfrac{\Sigma f(x - \bar{x})^2}{\Sigma f - 1} \\[5ex] \color{red}{Second\:\:Formula} \\[3ex] (5.)\:\: s^2 = \dfrac{n(\Sigma x^2) - (\Sigma x)^2}{n(n - 1)} \\[5ex] (6.)\:\: s^2 = \dfrac{\Sigma f(\Sigma fx^2) - (\Sigma fx)^2}{\Sigma f(\Sigma f - 1)} \\[7ex] \underline{Using\;\;Assumed\;\;Mean} \\[3ex] (7.)\;\; s^2 = \dfrac{\Sigma D^2}{n - 1} - \left(\dfrac{\Sigma D}{n - 1}\right)^2 \\[7ex] (8.)\;\; s^2 = \dfrac{\Sigma fD^2}{\Sigma f - 1} - \left(\dfrac{\Sigma fD}{\Sigma f - 1}\right)^2 \\[10ex] \underline{Population\:\:Variance} \\[3ex] \color{red}{First\:\:Formula} \\[3ex] (9.)\:\: \sigma^2 = \dfrac{\Sigma(x - \mu)^2}{N} \\[5ex] (10.)\:\: \sigma^2 = \dfrac{\Sigma f(x - \mu)^2}{\Sigma f} \\[5ex] \color{red}{Second\:\:Formula} \\[3ex] (11.)\:\: \sigma^2 = \dfrac{N(\Sigma x^2) - (\Sigma x)^2}{N^2} \\[5ex] (12.)\:\: \sigma^2 = \dfrac{\Sigma f(\Sigma fx^2) - (\Sigma fx)^2}{(\Sigma f)^2} \\[7ex] \underline{Using\;\;Assumed\;\;Mean} \\[3ex] (13.)\;\; \sigma^2 = \dfrac{\Sigma D^2}{N} - \left(\dfrac{\Sigma D}{N}\right)^2 \\[7ex] (14.)\;\; \sigma^2 = \dfrac{\Sigma fD^2}{\Sigma f} - \left(\dfrac{\Sigma fD}{\Sigma f}\right)^2 \\[10ex] \underline{Sample\:\:Standard\:\:Deviation} \\[3ex] \color{red}{First\:\:Formula} \\[3ex] (15.)\:\: s = \sqrt{\dfrac{\Sigma(x - \bar{x})^2}{n - 1}} \\[5ex] (16.)\:\: s = \sqrt{\dfrac{\Sigma f(x - \bar{x})^2}{\Sigma f - 1}} \\[5ex] \color{red}{Second\:\:Formula} \\[3ex] (17.)\:\: s = \sqrt{\dfrac{n(\Sigma x^2) - (\Sigma x)^2}{n(n - 1)}} \\[5ex] (18.)\:\: s = \sqrt{\dfrac{\Sigma f(\Sigma fx^2) - (\Sigma fx)^2}{\Sigma f(\Sigma f - 1)}} \\[7ex] \underline{Using\;\;Assumed\;\;Mean} \\[3ex] (19.)\;\; s = \sqrt{\dfrac{\Sigma D^2}{n - 1} - \left(\dfrac{\Sigma D}{n - 1}\right)^2} \\[7ex] (20.)\;\; s = \sqrt{\dfrac{\Sigma fD^2}{\Sigma f - 1} - \left(\dfrac{\Sigma fD}{\Sigma f - 1}\right)^2} \\[10ex] \underline{Population\:\:Standard\:\:Deviation} \\[3ex] \color{red}{First\:\:Formula} \\[3ex] (21.)\:\: \sigma = \sqrt{\dfrac{\Sigma(x - \mu)^2}{N}} \\[5ex] (22.)\:\: \sigma = \sqrt{\dfrac{\Sigma f(x - \mu)^2}{\Sigma f}} \\[5ex] \color{red}{Second\:\:Formula} \\[3ex] (23.)\:\: \sigma = \dfrac{\sqrt{N(\Sigma x^2) - (\Sigma x)^2}}{N} \\[5ex] (24.)\:\: \sigma = \dfrac{\sqrt{\Sigma f(\Sigma fx^2) - (\Sigma fx)^2}}{\Sigma f} \\[7ex] \underline{Using\;\;Assumed\;\;Mean} \\[3ex] (25.)\;\; \sigma = \sqrt{\dfrac{\Sigma D^2}{N} - \left(\dfrac{\Sigma D}{N}\right)^2} \\[7ex] (26.)\;\; \sigma = \sqrt{\dfrac{\Sigma fD^2}{\Sigma f} - \left(\dfrac{\Sigma fD}{\Sigma f}\right)^2} \\[10ex] \underline{Range\:\:Rule\:\:of\:\:Thumb} \\[3ex] Approximate\:\:Value\:\:of\:\:Calculating\:\:Standard\:\:Deviation \\[3ex] (27.)\:\: s = \dfrac{Range}{4} = \dfrac{max - min}{4} \\[7ex] \underline{Interquartile\:\:Range} \\[3ex] (28.)\:\: IQR = Q_3 - Q_1 \\[5ex] \underline{Coefficient\:\:of\:\:Variation\:\:for\:\:Sample} \\[3ex] (29.)\:\: CV = \dfrac{s}{x} * 100 ...in\:\:\% \\[7ex] \underline{Coefficient\:\:of\:\:Variation\:\:for\:\:Population} \\[3ex] (30.)\:\: CV = \dfrac{\sigma}{x} * 100 ...in\:\:\% \\[7ex] \underline{Mean\:\:Absolute\:\:Deviation} \\[3ex] (31.)\:\: MAD = \dfrac{\Sigma |x - \bar{x}|}{n} \\[5ex] \underline{Mean\:\:Absolute\:\:Deviation} \\[3ex] (32.)\:\: MAD = \dfrac{\Sigma f|x - \bar{x}|}{\Sigma f} \\[5ex] $


Measures of Spread: Grouped Data

$ \underline{Class\:\:Midpoint} \\[3ex] (1.)\:\: x_{mid} = \dfrac{LCL + UCL}{2} \\[5ex] \underline{Using\;\;Assumed\;\;Mean} \\[3ex] (2.)\;\; D = x_{mid} - AM \\[5ex] \underline{Sample\:\:Variance} \\[3ex] \color{red}{First\:\:Formula} \\[3ex] (3.)\:\: s^2 = \dfrac{\Sigma f(x_{mid} - \bar{x})^2}{\Sigma f - 1} \\[5ex] \color{red}{Second\:\:Formula} \\[3ex] (4.)\:\: s^2 = \dfrac{\Sigma f(\Sigma fx_{mid}^2) - (\Sigma fx_{mid})^2}{\Sigma f(\Sigma f - 1)} \\[5ex] \underline{Using\;\;Assumed\;\;Mean} \\[3ex] (5.)\;\; s^2 = \dfrac{\Sigma D^2}{n - 1} - \left(\dfrac{\Sigma D}{n - 1}\right)^2 \\[7ex] (6.)\;\; s^2 = \dfrac{\Sigma fD^2}{\Sigma f - 1} - \left(\dfrac{\Sigma fD}{\Sigma f - 1}\right)^2 \\[10ex] \underline{Sample\:\:Standard\:\:Deviation} \\[3ex] \color{red}{First\:\:Formula} \\[3ex] (7.)\:\: s = \sqrt{\dfrac{\Sigma f(x_{mid} - \bar{x})^2}{\Sigma f - 1}} \\[5ex] \color{red}{Second\:\:Formula} \\[3ex] (8.)\:\: s = \sqrt{\dfrac{\Sigma f(\Sigma fx_{mid}^2) - (\Sigma fx_{mid})^2}{\Sigma f(\Sigma f - 1)}} \\[5ex] \underline{Using\;\;Assumed\;\;Mean} \\[3ex] (9.)\;\; s = \sqrt{\dfrac{\Sigma D^2}{n} - \left(\dfrac{\Sigma D}{n - 1}\right)^2} \\[7ex] (10.)\;\; s = \sqrt{\dfrac{\Sigma fD^2}{\Sigma f - 1} - \left(\dfrac{\Sigma fD}{\Sigma f - 1}\right)^2} \\[10ex] \underline{Population\:\:Variance} \\[3ex] \color{red}{First\:\:Formula} \\[3ex] (11.)\:\: \sigma^2 = \dfrac{\Sigma f(x_{mid} - \bar{x})^2}{\Sigma f} \\[5ex] \color{red}{Second\:\:Formula} \\[3ex] (12.)\:\: \sigma^2 = \dfrac{\Sigma f(\Sigma fx_{mid}^2) - (\Sigma fx_{mid})^2}{\Sigma f(\Sigma f)} \\[5ex] \underline{Using\;\;Assumed\;\;Mean} \\[3ex] (13.)\;\; \sigma^2 = \dfrac{\Sigma D^2}{N} - \left(\dfrac{\Sigma D}{N}\right)^2 \\[7ex] (14.)\;\; \sigma^2 = \dfrac{\Sigma fD^2}{\Sigma f} - \left(\dfrac{\Sigma fD}{\Sigma f}\right)^2 \\[10ex] \underline{Population\:\:Standard\:\:Deviation} \\[3ex] \color{red}{First\:\:Formula} \\[3ex] (15.)\:\: \sigma = \sqrt{\dfrac{\Sigma f(x_{mid} - \bar{x})^2}{\Sigma f}} \\[5ex] \color{red}{Second\:\:Formula} \\[3ex] (16.)\:\: \sigma = \sqrt{\dfrac{\Sigma f(\Sigma fx_{mid}^2) - (\Sigma fx_{mid})^2}{\Sigma f(\Sigma f)}} \\[5ex] \underline{Using\;\;Assumed\;\;Mean} \\[3ex] (17.)\;\; \sigma = \sqrt{\dfrac{\Sigma D^2}{N} - \left(\dfrac{\Sigma D}{N}\right)^2} \\[7ex] (18.)\;\; \sigma = \sqrt{\dfrac{\Sigma fD^2}{\Sigma f} - \left(\dfrac{\Sigma fD}{\Sigma f}\right)^2} \\[10ex] $


Measures of Position

A data value is usual if $-2.00 \le z-score \le 2.00$

A data value is unusual if $z-score \lt -2.00$ OR $z-score \gt 2.00$

$ \underline{Sample} \\[3ex] Minimum\:\:usual\:\:data\:\:value = \bar{x} - 2s \\[3ex] Maximum\:\:usual\:\:data\:\:value = \bar{x} + 2s \\[5ex] \underline{Population} \\[3ex] Minimum\:\:usual\:\:data\:\:value = \mu - 2\sigma \\[3ex] Maximum\:\:usual\:\:data\:\:value = \mu + 2\sigma \\[5ex] \underline{z\:\:score\:\:for\:\:Sample} \\[3ex] (1.)\:\: z = \dfrac{x - \bar{x}}{s} \\[7ex] \underline{z\:\:score\:\:for\:\:Population} \\[3ex] (2.)\:\: z = \dfrac{x - \mu}{\sigma} \\[7ex] \underline{Quantiles(Percentiles,\:Deciles,\:Quintiles,\:and\:Quartiles)} \\[3ex] \color{red}{Convert\:\:a\:\:Data\:\:value\:\:to\:\:a\:\:Quantile} \\[3ex] x\:\:and\:\:y\:\:are\:\:two\:\:different\:\:variables \\[3ex] (3.)\:\: Percentile\:\:of\:\:x = \dfrac{number\:\:of\:\:values\:\:less\:\:than\:\:x}{total\:\:number\:\:of\:\:values} * 100 = yth\:\:Percentile \\[5ex] (4.)\:\: Decile\:\:of\:\:x = \dfrac{number\:\:of\:\:values\:\:less\:\:than\:\:x}{total\:\:number\:\:of\:\:values} * 10 = yth\:\:Decile \\[5ex] (5.)\:\: Quintile\:\:of\:\:x = \dfrac{number\:\:of\:\:values\:\:less\:\:than\:\:x}{total\:\:number\:\:of\:\:values} * 5 = yth\:\:Quintile \\[5ex] (6.)\:\: Quartile\:\:of\:\:x = \dfrac{number\:\:of\:\:values\:\:less\:\:than\:\:x}{total\:\:number\:\:of\:\:values} * 4 = yth\:\:Quartile \\[7ex] \color{red}{Convert\:\:a\:\:Quantile\:\:to\:\:a\:\:Data\:\:Value} \\[3ex] Calculate\:\:the\:\:xth\:\:position\:\:of\:\:the\:\:yth\:\:Quantile \\[3ex] (7.)\:\: xth\:\:position = \dfrac{yth\:\:Percentile}{100} * total\:\:number\:\:of\:\:values \\[5ex] (8.)\:\: xth\:\:position = \dfrac{yth\:\:Decile}{10} * total\:\:number\:\:of\:\:values \\[5ex] (9.)\:\: xth\:\:position = \dfrac{yth\:\:Quintile}{5} * total\:\:number\:\:of\:\:values \\[5ex] (10.)\:\: xth\:\:position = \dfrac{yth\:\:Quartile}{4} * total\:\:number\:\:of\:\:values \\[7ex] $


If the $xth$ position then,
is an integer
$xth\:\:position = \dfrac{xth\:\:position + (x + 1)th\:\;position}{2}$

In other words, find the value of the $xth$ position; find the value of the next position; and determine the mean of the two values.
is not an integer $xth$ position is rounded up


$ \underline{The\:\:Five-Number\:\:Summary\:\:of\:\:Data} \\[3ex] (11.)\:\: Minimum\:(min) \\[3ex] (12.)\:\: Lower\:\:Quartile\:(Q_1) \\[3ex] (13.)\:\: Median\:\:or\:\:Middle\:\:Quartile\:(Q_2) \\[3ex] (14.)\:\: Upper\:\:Quartile\:(Q_3) \\[3ex] (15.)\:\: Maximum\:(Max) \\[5ex] \underline{Other\:\:Statistics\:\:from\:\:Quantiles} \\[3ex] (16.)\:\: IQR = Q_3 - Q_1 \\[3ex] (17.)\:\: SIQR = \dfrac{IQR}{2} = \dfrac{Q_3 - Q_1}{2} \\[5ex] (18.)\:\: MQ = \dfrac{Q_3 + Q_1}{2} \\[5ex] (19.)\:\: Upper\:\:Quartile\:(Q_3) \\[3ex] (20.)\:\: LF = Q_1 - 1.5(IQR) \\[3ex] (21.)\:\: UF = Q_3 + 1.5(IQR) $


Probability

Given any two events say A and B

$ P(E) = \dfrac{n(E)}{n(S)} \\[5ex] \underline{\text{Addition Rule}} \\[3ex] \dfrac{n(A \cup B)}{n(S)} = \dfrac{n(A)}{n(S)} + \dfrac{n(B)}{n(S)} - \dfrac{n(A \cap B)}{n(S)} \\[5ex] P(A \cup B) = P(A) + P(B) - P(A \cap B) \\[3ex] P(A\:\:\:OR\:\:\:B) = P(A) + P(B) - P(A\:\:\:AND\:\:\:B) \\[5ex] $ For Independent Events

$ P(B|A) = P(B) \\[3ex] \rightarrow P(A\:\:\:OR\:\:\:B) = P(A) + P(B) - [P(A) * P(B)] \\[5ex] $ For Dependent Events

$ P(B|A) = P(B|A) \\[3ex] \rightarrow P(A\:\:\:OR\:\:\:B) = P(A) + P(B) - [P(A) * P(B|A)] \\[5ex] $ For Mutually Exclusive Events (Disjoint Events)

$ P(A \cap B) = 0 \\[3ex] P(A\:\:\:OR\:\:\:B) = P(A) + P(B) - 0 \\[3ex] \rightarrow P(A\:\:\:OR\:\:\:B) = P(A) + P(B) \\[5ex] $
$ \underline{\text{Multiplication Rule}} \\[3ex] P(A\:\:\:AND\:\:\:B) = P(A) * P(B|A) \\[3ex] P(A \cap B) = P(A) * P(B|A) \\[3ex] P(A\:\:\:AND\:\:\:B) = P(A \cap B) \\[5ex] $ $P(B|A)$ is read as: the probability of event $B$ given event $A$

For Independent Events

$ P(B|A) = P(B) \\[3ex] \rightarrow P(A\:\:\:AND\:\:\:B) = P(A) * P(B) \\[5ex] $ For Dependent Events

$ P(B|A) = P(B|A) \\[3ex] \rightarrow P(A\:\:\:AND\:\:\:B) = P(A) * P(B|A) \\[5ex] $ The complement of Event $A$ is $A'$

$ \underline{Complementary\;\;Rule} \\[3ex] P(A) + P(A') = 1 \\[3ex] \rightarrow P(A') = 1 - P(A) \\[5ex] $ Other Formulas

$ (1.)\;\; P(A) = P(A \cap B') + P(A \cap B) $


Probability Distributions

$ \boldsymbol{Probability\;\;Distribution} \\[3ex] (1.)\;\;\mu = \Sigma[x * P(x)] \\[3ex] (2.)\;\;E = \Sigma[x * P(x)] \\[3ex] (3.)\;\; \sigma = \sqrt{\Sigma[x^2 * P(x)] - \mu^2} \\[7ex] \boldsymbol{Combinatorics} \\[3ex] (1.)\:\: 0! = 1 \\[3ex] (2.)\:\: n! = n * (n - 1) * (n - 2) * (n - 3) * ... * 1 \\[3ex] (3.)\;\; n! = n * (n - 1)! \\[3ex] (4.)\;\; n! = (n - 1) * (n - 2)!...among\;\;others \\[3ex] (5.)\:\: C(n, x) = \dfrac{n!}{(n - x)!x!} \\[5ex] (6.)\;\; C(n, x) = C(n, n - x) \\[7ex] \boldsymbol{Binomial\;\;Distribution} \\[3ex] (1.)\;\; p + q = 1 \\[3ex] (2.)\;\; \mu = n * p \\[3ex] (3.)\;\; \sigma = \sqrt{n * p * q} \\[4ex] (4.)\;\; P(x) = C(n, x) * p^x * q^{n - x}...\text{Depends on the context of the question} \\[5ex] where \\[3ex] x = \text{number of successes/failures} \\[3ex] n = \text{number of trials} = 12 \\[3ex] C(n, x) = \text{Binomial coefficient} \\[3ex] P(x) = \text{Probability of the number of successes/failures} \\[3ex] p = \text{probability of success} = 70\% = 0.7 \\[3ex] q = \text{probability of failure} = 1 - 0.7 = 0.3 \\[5ex] \boldsymbol{Poisson\;\;Distribution} \\[3ex] (1.)\;\;P(x) = \dfrac{\mu^x * e^{-\mu}}{x!} \\[5ex] (2.)\;\; \mu = \sigma^2 \\[7ex] \boldsymbol{Normal\;\;Distribution} \\[3ex] (1.)\;\; z = \dfrac{x - \bar{x}}{s} \\[5ex] (2.)\;\; x = \bar{x} + zs \\[3ex] (3.)\;\; z = \dfrac{x - \mu}{\sigma} \\[5ex] (4.)\;\; x = \mu + z\sigma \\[3ex] (5.)\;\;\text{Probability Density Function},\;\;P(x) = \dfrac{1}{\sigma\sqrt{2\pi}}e^{{-\dfrac{1}{2}}\left(\dfrac{x - \mu}{\sigma}\right)^2} \\[7ex] $

Empirical Rule (68 - 95 - 99.7 percent Rule)
(Applies only to Normal Distribution)
(a.) 68% of the data lie within (below and above) 1 standard deviation of the mean
(b.) 95% of the data lie within (below and above) 2 standard deviations of the mean
(c.) 99.7% of the data lie within (below and above) 3 standard deviations of the mean

Pafnuty Chebyshev's Theorem
(Applies to any distribution)
At least $\left(1 - \dfrac{1}{k^2}\right) * 100$ % of the data lie within $k$ standard deviations of the mean
implies
At least $\left(1 - \dfrac{1}{k^2}\right) * 100$ % of the data lie within $\mu - k\sigma$ and $\mu + k\sigma$

Range Rule of Thumb
Minimum Usual Value = μ - 2σ
Maximum Usual Value = μ + 2σ
A data value is unusual if it is less than the minimum usual value or greater than the maximum usual value

z-score Boundary
A data value is usual if −2.00 ≤ z-score ≤ 2.00
A data value is unusual if z-score < −2.00 or if z-score > 2.00

(1.) The table shows the distribution of distance (in km) of 60 villages from a state capital.

Distance (in km) 0 – 19 20 – 29 30 – 39 40 – 49 50 – 69 70 – 99 100 – 149
Number of villages 12 7 6 8 5 9 10

Draw a histogram for the distribution.


$ \text{Lower Class Boundary} = \text{Lower Class Interval} - 0.5...\text{because of integers} \\[3ex] \text{Upper Class Boundary} = \text{Upper Class Interval} + 0.5...\text{because of integers} \\[3ex] \text{Class Width} = \text{Upper Class Boundary} - \text{Lower Class Boundary} \\[3ex] \text{Frequency Density} = \dfrac{\text{Frequency}}{\text{Class Width}} \\[5ex] $
Class Intervals: Distance (km) Class Boundaries Number of Villages: Frequency, F Class Width Frequency Density
0 – 19 −0.5 – 19.5 12 20 0.6
20 – 29 19.5 – 29.5 7 10 0.7
30 – 39 29.5 – 39.5 6 10 0.6
40 – 49 39.5 – 49.5 8 10 0.8
50 – 69 49.5 – 69.5 5 20 0.25
70 – 99 69.5 – 99.5 9 30 0.3
100 – 149 99.5 – 149.5 10 50 0.2

Scale:
1 cm to 0.1 unit on the y-axis (Frequency Density axis)
2 cm to 20 units on the x-axis (Distance axis)

The histogram is drawn as shown:

Number 1
(2.)


(3.) The data shows the ordered marks scored by students in a test: $11, 12, (2x + y), (x + 2y), 14$ and $(y^2 - 2x)$.
Given that the median is $13\dfrac{1}{2}$ and y is greater than x by 1, find:
(a.) the values of x and y;
(b.) correct to three significant figures, the standard deviation of the distribution.


The dataset is ordered

$ 11, 12, (2x + y), (x + 2y), 14, (y^2 - 2x) \\[3ex] y = x + 1...eqn.(1) ...\text{greater than }x \;\;by\;\;1 \\[3ex] Median = 13\dfrac{1}{2} = \dfrac{27}{2} \\[5ex] Median = \dfrac{(2x + y) + (x + 2y)}{2} \\[5ex] \implies \\[3ex] \dfrac{(2x + y) + (x + 2y)}{2} = \dfrac{27}{2} \\[5ex] \text{same denominator; equate the numerators} \\[3ex] 2x + y + x + 2y = 27 \\[3ex] 3x + 3y = 27 \\[3ex] 3(x + y) = 27 \\[3ex] x + y = \dfrac{27}{3} \\[5ex] x + y = 9...eqn.(2) \\[3ex] \text{Substitute for y in eqn.(2)} \\[3ex] x + x + 1 = 9 \\[3ex] 2x = 9 - 1 \\[3ex] 2x = 8 \\[3ex] x = \dfrac{8}{2} \\[5ex] x = 4 \\[3ex] \text{Substitute for x in eqn.(1)} \\[3ex] y = 4 + 1 \\[3ex] y = 5 \\[5ex] 2x + y = 2(4) + 5 = 13 \\[3ex] x + 2y = 4 + 2(5) = 14 \\[3ex] y^2 - 2x = 5^2 - 2(4) = 17 \\[3ex] \text{Dataset is: } 11, 12, 13, 14, 14, 17 \\[3ex] \text{Size}, n = 6 \\[3ex] $ We shall assume that the dataset is a population
Let us set up the table to do the calculations.

$x$ $x - \bar{x}$ (x - \bar{x})^2
11 −2.5 6.25
12 −1.5 2.25
13 −0.5 0.25
14 0.5 0.25
14 0.5 0.25
17 3.5 12.25
$\Sigma x = 81$ $ Mean, \bar{x} = \dfrac{\Sigma x}{n} = \dfrac{81}{6} \\[5ex] = 13.5 $ $\Sigma (x - \bar{x})^2 = 21.5$

$ \text{Standard Deviation}, \mu = \sqrt{\dfrac{\Sigma (x - \bar{x})^2}{n}} \\[5ex] = \sqrt{\dfrac{21.5}{6}} \\[5ex] = 1.892969449 \\[3ex] \approx 1.89 ...\text{to 3 significant figures} $

Calculator 3a

Calculator 3b

Calculator 3c
(4.) Two events M and N are such that $P(M) = \dfrac{1}{2}$, $P(N) = \dfrac{9}{20}$ and $P(M \cap N) = \dfrac{11}{50}$.

Find:
(a.) $P(M \cap N')$
(b.) $P(M' \cap N)$


$ (a.) \\[3ex] P(M \cap N') = P(M\;\;only) \\[3ex] P(M) = P(M \cap N') + P(M \cap N) \\[3ex] P(M \cap N') = P(M) - P(M \cap N) \\[3ex] = \dfrac{1}{2} - \dfrac{11}{50} \\[5ex] = \dfrac{25 - 11}{50} \\[5ex] = \dfrac{14}{50} \\[5ex] = \dfrac{7}{25} \\[5ex] (b.) \\[3ex] P(M' \cap N) = P(N\;\;only) \\[3ex] P(N) = P(M' \cap N) + P(M \cap N) \\[3ex] P(M' \cap N) = P(N) - P(M \cap N) \\[3ex] = \dfrac{9}{20} - \dfrac{11}{50} \\[5ex] = \dfrac{45 - 22}{100} \\[5ex] = \dfrac{23}{100} $
(5.) In an examination, 70% of the candidates passed.
If 12 candidates are selected at random, find the probability that:
(a.) at least two of them failed;
(b.) exactly half of them passed;
(c.) not more than one-sixth of them failed.


$ \underline{\text{Binomial Probability Distribution}} \\[3ex] P(x) = C(n, x) * p^x * q^{n - x}...\text{depending on the context} \\[4ex] where \\[3ex] x = \text{number of successes/failures} ...\text{depending on the context} \\[3ex] n = \text{number of trials} \\[3ex] C(n, x) = \text{Binomial coefficient} \\[3ex] P(x) = \text{Probability of the number of successes/failures} \\[3ex] p = \text{probability of success} \\[3ex] q = \text{probability of failure} \\[5ex] (a.) \\[3ex] \text{at least 2 failed} \\[3ex] x \ge 2 \\[3ex] n = 12 \\[3ex] P(x) = ? \\[3ex] p = 70\% = 0.7 \\[3ex] q = 1 - 0.7 = 0.3 \\[5ex] P(x \ge 2) + P(x \lt 2) = 1 ...\text{Complementary Rule} \\[3ex] P(x \ge 2) = 1 - P(x \lt 2) \\[3ex] = 1 - [P(x = 0) + P(x = 1)] \\[5ex] P(x) = C(n, x) * q^x * p^{n - x}...\text{in the context of failing} \\[4ex] P(x = 0) = C(12, 0) * (0.3)^0 * (0.7)^{12 - 0} \\[4ex] = \dfrac{12!}{(12 - 0)! * 0!} * 1 * 0.7^{12} \\[5ex] = \dfrac{12!}{12! * 1} * 0.0138412872 \\[5ex] = 0.0138412872 \\[5ex] P(x = 1) = C(12, 1) * (0.3)^1 * (0.7)^{12 - 1} \\[4ex] = \dfrac{12!}{(12 - 1)! * 1!} * 0.3 * 0.7^{11} \\[5ex] = \dfrac{12 * 11!}{11! * 1} * 0.3 * 0.0197732674 \\[5ex] = 0.0711837627 \\[5ex] \implies \\[3ex] P(x \ge 2) = 1 - (0.0138412872 + 0.0711837627) \\[3ex] = 1 - 0.0850250499 \\[3ex] = 0.9149749501 \\[5ex] (b.) \\[3ex] \text{half of them} = \dfrac{1}{2} * 12 = 6 \\[5ex] P(x) = C(n, x) * p^x * q^{n - x}...\text{in the context of passing} \\[4ex] P(x = 6) = C(12, 6) * 0.7^6 * 0.3^{12 - 6} \\[4ex] = \dfrac{12!}{(12 - 6)! * 6!} * 0.117649 * 0.3^6 \\[5ex] = \dfrac{12!}{6! * 6!} * 0.117649 * 0.000729 \\[5ex] = 0.0792478958 \\[5ex] (c.) \\[3ex] \text{one-sixth of them} = \dfrac{1}{6} * 12 = 2 \\[5ex] \text{not more than 2 failed} = \text{at most 2 failed} \\[3ex] x \le 2 \\[3ex] P(x \le 2) = P(x = 0) + P(x = 1) + P(x = 2) \\[3ex] P(x) = C(n, x) * q^x * p^{n - x}...\text{in the context of failing} \\[4ex] P(x = 2) = C(12, 2) * 0.3^2 * 0.7^{12 - 2} \\[4ex] = \dfrac{12!}{(12 - 2)! * 2!} * 0.09 * 0.7^{10} \\[5ex] = \dfrac{12!}{10! * 2!} * 0.09 * 0.0282475249 \\[5ex] = 0.1677902979 \\[3ex] \therefore P(x \le 2) = 0.0138412872 + 0.0711837627 + 0.1677902979 \\[3ex] = 0.2528153479 $

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