Understanding the Mean and Standard Deviation

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Question 1 of 2

The height of a group of

5

$5$ dogs and a group of

5

$5$ cats are collected in centimetres. Compare the mean for dogs

¯ x

$bar x$ _D and the mean for cats

¯ x

$bar x$ _C.

$¯ x$ $bar x$ _D $=$ $=$ (40) $c m$ $cm$

$¯ x$ $bar x$ _C $=$ $=$ (40) $c m$ $cm$

Question 2 of 2

The height of a group of

5

$5$ dogs and a group of

5

$5$ cats are collected in centimetres. Find and compare the standard deviation of both groups.

The mean of both groups is given:

¯ ¯ ¯ X = 40

$barX=40$

$S_{d} =$ $S_d=$ (14.1) $c m$ $cm$

$S_{c} =$ $S_c=$ (1.4) $c m$ $cm$

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Formula for Standard Deviation

S = \sqrt{\frac{\sum (X - {¯ X)}^{2}}{n}}

$S=\sqrt{\frac{\sum(\color{#9a00c7}{X}-\color{#00880a}{\bar{X)}}^2}{\color{#004ec4}{n}}}$

To solve for the standard deviation for the height of the dogs, label the given values needed for the formula.

X values (heights) = 20, 30, 40, 50, 60

$X \text(values) \text((heights))=20,30,40,50,60$

$Mean ¯ ¯ ¯ X = 40$ $\text(Mean) barX=40$

$Number of dogs : n = 5$ $\text(Number of dogs): n=5$

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

	$=$ $=$	$\sqrt{\frac{{(- 20)}^{2} + {(- 10)}^{2} + {(0)}^{2} + {(10)}^{2} + {(20)}^{2}}{5}}$ $sqrt(((-20)^2+(-10)^2+(0)^2+(10)^2+(20)^2)/5)$

	$=$ $=$	$\sqrt{\frac{400 + 100 + 0 + 100 + 400}{5}}$ $sqrt((400+100+0+100+400)/5)$

	$=$ $=$	$\sqrt{\frac{1000}{5}}$ $sqrt(1000/5)$

	$=$ $=$	$\sqrt{200}$ $sqrt(200)$
$S_{d}$ $S_d$	$=$ $=$	$14.1$ $14.1$

To solve for the standard deviation for the height of the cats, label the given values needed for the formula.

X values (heights) = 38, 39, 40, 41, 42

$X \text(values) \text((heights))=38,39,40,41,42$

$Mean ¯ ¯ ¯ X = 40$ $\text(Mean) barX=40$

$Number of cats : n = 5$ $\text(Number of cats): n=5$

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

	$=$ $=$	$\sqrt{\frac{{(- 2)}^{2} + {(- 1)}^{2} + {(0)}^{2} + {(1)}^{2} + {(2)}^{2}}{5}}$ $sqrt(((-2)^2+(-1)^2+(0)^2+(1)^2+(2)^2)/5)$

	$=$ $=$	$\sqrt{\frac{4 + 1 + 0 + 1 + 4}{5}}$ $sqrt((4+1+0+1+4)/5)$

	$=$ $=$	$\sqrt{\frac{10}{5}}$ $sqrt(10/5)$	Simplify

	$=$ $=$	$\sqrt{2}$ $sqrt2$
$S_{c}$ $S_c$	$=$ $=$	$1.4$ $1.4$

S_{d}

$S_d$ is greater than

S_{c}

$S_c$ .

This means that the height of the dogs is more dispersed than the height of the cats.

S_{d} = 14.1 c m

$S_d=14.1cm$

S_{c} = 1.4 c m

$S_c=1.4cm$

Understanding the Mean and Standard Deviation

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Quiz summary

Information

1. Question

Hint

Fantastic!

Formula for Mean

2. Question

Hint

Keep Going!

Formula for Standard Deviation

Quizzes

$¯ x$ $bar x$ _D	$=$ $=$	$\frac{sum of data}{number of data}$ $(\text(sum of data))/(\text(number of data))$

	$=$ $=$	$\frac{20 + 30 + 40 + 50 + 60}{5}$ $(20+30+40+50+60)/5$	Substitute values

	$=$ $=$	$\frac{200}{5}$ $200/5$

	$=$ $=$	$40$ $40$ $c m$ $cm$	Simplify

$¯ x$ $bar x$ _C	$=$ $=$	$\frac{sum of data}{number of data}$ $(\text(sum of data))/(\text(number of data))$

	$=$ $=$	$\frac{38 + 39 + 40 + 41 + 42}{5}$ $(38+39+40+41+42)/5$	Substitute values

	$=$ $=$	$\frac{200}{5}$ $200/5$

	$=$ $=$	$40$ $40$ $c m$ $cm$	Simplify