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Mean and Standard DeviationMean and Standard Deviation
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Question 1 of 2
1. Question
The height of a group of `5` dogs and a group of `5` cats are collected in centimetres. Compare the mean for dogs `bar x`D and the mean for cats `bar x`C.
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`bar x`D`=` (40)`cm``bar x`C`=` (40)`cm`
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Formula for Mean
`bar x=(\text(sum of data))/(\text(number of data))``bar x`D`=` dogs mean`bar x`C`=` cats meanUse the mean formula to find the average height of the dogs.
`bar x`D `=` `(\text(sum of data))/(\text(number of data))` `=` `(20+30+40+50+60)/5` Substitute values `=` `200/5` `=` `40` `cm` Simplify Use the mean formula to find the average height of the cats.
`bar x`C `=` `(\text(sum of data))/(\text(number of data))` `=` `(38+39+40+41+42)/5` Substitute values `=` `200/5` `=` `40` `cm` Simplify The mean of both groups is `40` cm even though their heights are not the same.This only means that both groups have their data centralized or averaged at `40` cm.`bar x`D`=40` `cm``bar x`C`=40` `cm` -
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Question 2 of 2
2. Question
The height of a group of `5` dogs and a group of `5` cats are collected in centimetres. Find and compare the standard deviation of both groups.The mean of both groups is given: `barX=40`
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`S_d=` (14.1) `cm``S_c=` (1.4) `cm`
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Formula for Standard Deviation
$$S=\sqrt{\frac{\sum(\color{#9a00c7}{X}-\color{#00880a}{\bar{X)}}^2}{\color{#004ec4}{n}}}$$To solve for the standard deviation for the height of the dogs, label the given values needed for the formula.
`X \text(values) \text((heights))=20,30,40,50,60``\text(Mean) barX=40``\text(Number of dogs): n=5`Subsitute into the standard deviation formula$$S$$ `=` $$\sqrt{\frac{\sum(\color{#9a00c7}{X}-\color{#00880a}{\bar{X)}}^2}{\color{#004ec4}{n}}}$$ `=` $$\sqrt{\frac{(\color{#9a00c7}{20}-\color{#00880a}{40})^2+(\color{#9a00c7}{30}-\color{#00880a}{40})^2+(\color{#9a00c7}{40}-\color{#00880a}{40})^2+(\color{#9a00c7}{50}-\color{#00880a}{40})^2+(\color{#9a00c7}{60}-\color{#00880a}{40})^2}{\color{#004ec4}5}}$$ Substitute values `=` `sqrt(((-20)^2+(-10)^2+(0)^2+(10)^2+(20)^2)/5)` `=` `sqrt((400+100+0+100+400)/5)` `=` `sqrt(1000/5)` `=` `sqrt(200)` `S_d` `=` `14.1` To solve for the standard deviation for the height of the cats, label the given values needed for the formula.
`X \text(values) \text((heights))=38,39,40,41,42``\text(Mean) barX=40``\text(Number of cats): n=5`Subsitute into the standard deviation formula$$S$$ `=` $$\sqrt{\frac{\sum(\color{#9a00c7}{X}-\color{#00880a}{\bar{X)}}^2}{\color{#004ec4}{n}}}$$ `=` $$\sqrt{\frac{(\color{#9a00c7}{38}-\color{#00880a}{40})^2+(\color{#9a00c7}{39}-\color{#00880a}{40})^2+(\color{#9a00c7}{40}-\color{#00880a}{40})^2+(\color{#9a00c7}{41}-\color{#00880a}{40})^2+(\color{#9a00c7}{42}-\color{#00880a}{40})^2)}{\color{#004ec4}5}}$$ Substitute values `=` `sqrt(((-2)^2+(-1)^2+(0)^2+(1)^2+(2)^2)/5)` `=` `sqrt((4+1+0+1+4)/5)` `=` `sqrt(10/5)` Simplify `=` `sqrt2` `S_c` `=` `1.4` `S_d` is greater than `S_c`.This means that the height of the dogs is more dispersed than the height of the cats.`S_d=14.1cm``S_c=1.4cm` -