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Statistics and Probability

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These are the solutions to the Cambridge O-Level (Ordinary Level) past questions on Statistics and Probability.
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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.) A shop sells two varieties of apple tree.
(a.) The cumulative frequency diagram shows the heights, in metres, of 80 Variety A trees.

Number 1

(i.) Use the diagram to estimate
(a.) the median
(b.) the 30th percentile.

(ii.) Trees with a height greater than ym are graded Class I.
$\dfrac{2}{5}$ of the 80 trees are graded Class I.
Find the value of y.

(iii.) Complete the frequency table for the heights of the Variety A trees.
Height (h m) 1.1 < h ≤ 1.2 1.2 < h ≤ 1.3 1.3 < h ≤ 1.4 1.4 < h ≤ 1.5 1.5 < h ≤ 1.6
Frequency 6 24

(b.) The frequency table shows the heaights of 50 Variety B trees.
Height (h m) 1.5 < h ≤ 1.7 1.7 < h ≤ 1.8 1.8 < h ≤ 1.9 1.9 < h ≤ 2.3
Frequency p 15 17 q

Using the midpoints of the intervals, the estimated mean height of these Variety B trees is 1.81 m.
Calculate the value of p and the value of q.


$ \underline{\text{Scales}} \\[3ex] \text{Cumulative Frequency axis}: \\[3ex] 5\;lines \rightarrow 10 - 0 \\[3ex] 5\;lines \rightarrow 10 \\[3ex] 1\;line \rightarrow \dfrac{10}{5} = 2\;units \\[5ex] \text{Height axis}: \\[3ex] 10\;lines \rightarrow 1.2 - 1.1 \\[3ex] 10\;lines \rightarrow 0.1 \\[3ex] 1\;line \rightarrow \dfrac{0.1}{10} = 0.01\;unit \\[5ex] (i.) \\[3ex] size = N = 80 \\[3ex] (a.) \\[3ex] \text{Median = 50th percentile} \\[3ex] = \left(\dfrac{50}{100} * 80\right)th \\[5ex] = 40th\;height \\[3ex] = 1.33\;m \\[5ex] (b.) \\[3ex] \text{30th percentile} \\[3ex] = \left(\dfrac{30}{100} * 80\right)th \\[5ex] = 24th\;height \\[3ex] = 1.28\;m \\[5ex] (ii.) \\[3ex] \text{Class I trees} = \dfrac{2}{5} * 80 \\[5ex] = 32\;trees \\[3ex] \text{To find } y: \\[3ex] 80 - 32 = 48 \\[3ex] 48 \rightarrow what\;\;height? \\[3ex] $ Number 1

$ y = 1.36\;m \\[3ex] $ (iii.)
Height (h m) 1.1 < h ≤ 1.2 1.2 < h ≤ 1.3 1.3 < h ≤ 1.4 1.4 < h ≤ 1.5 1.5 < h ≤ 1.6
Frequency 6 30 − 6 = 24 58 − 30 = 28 76 − 58 = 18 80 − 76 = 4

(b.)
Height, h(m) Midpoint, x (m) Frequency, F $F * x$
1.5 < h ≤ 1.7 $\dfrac{1.5 + 1.7}{2} = 1.6$ p 1.6p
1.7 < h ≤ 1.8 $\dfrac{1.7 + 1.8}{2} = 1.75$ 15 26.25
1.8 < h ≤ 1.9 $\dfrac{1.8 + 1.9}{2} = 1.85$ 17 31.45
1.9 < h ≤ 2.3 $\dfrac{1.9 + 2.3}{2} = 2.1$ q 2.1q
$ \Sigma F = p + q + 32 \\[3ex] \Sigma F = 50 $ $\Sigma Fx = 1.6p + 2.1q + 57.7$

$ p + q + 32 = 50 \\[3ex] p + q = 50 - 32 \\[3ex] p + q = 18 ...eqn.(1) \\[5ex] Mean, \bar{x} = \dfrac{\Sigma Fx}{\Sigma F} \\[5ex] 1.81 = \dfrac{1.6p + 2.1q + 57.7}{50} \\[5ex] 1.6p + 2.1q + 57.7 = 1.81(50) \\[3ex] 1.6p + 2.1q = 90.5 - 57.7 \\[3ex] 1.6p + 2.1q = 32.8 \\[3ex] 16p + 21q = 328 ...eqn.(2) \\[3ex] 16 * eqn.(1) \implies \\[3ex] 16p + 16q = 288 ...eqn.(3) \\[3ex] eqn.(2) - eqn.(3) \implies \\[3ex] 16p + 21q - (16p + 16q) = 328 - 288 \\[3ex] 16p + 21q - 16p - 16q = 40 \\[3ex] 5q = 40 \\[3ex] q = \dfrac{40}{5} \\[5ex] q = 8 \\[3ex] \text{Substitute for q in eqn.(1)} \\[3ex] p = 18 - q \\[3ex] p = 18 - 8 \\[3ex] p = 10 $
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