Question

3,2 The ollowing set of data is from a sample of n - 6: 74973 12 a. Compute the mean, median, and mode. b. Compute the range, variance, standard deviation, and coeffi- cient of variation. c. Compute the Z scores. Are there any outliers? d. Describe the shape of the data set.

Answer 1

Given , $n ~=~6$

and data are   $7,~4,~9,~7,~3,~12$

(a) Now, finding mean of the given data :

$\bar{X} ~=~ \frac{\sum_{x=1} ^{n}}{n}$

$\implies$  $\bar{X}$   $=$   $\frac{7 + 4+9+7+3+12 }{6} ~ = ~ 7$ .

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Now, about to median ;

Firstly arrange the data in an array : $3,~ 4,~ 7,~ 7,~ 9,~ 12$

Since there are even number of observations and $7,~ 7$ are the two middle terms of the data .

So, just take the arithmetic mean of both : i.e.

$\implies$$Median ~=~ \frac{7~+~7}{2} ~=~ 7$

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Now, proceeding to mode in the data : $3,~ 4,~ 7,~ 7,~ 9,~ 12$

Here, we can clearly see that $7$ is the observation that occurred most in the data .

Hence , $Mode ~=~7$

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(b)  $Range ~=~ Highest ~ value ~-~ Lowest ~ value$

$\implies$ here, our lowest value is 3 and highest value is 12 .

$\implies$$Range ~=~ 12 ~-~ 3 ~=~ 9$ .

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Now,   $Variance ~=~ \sum \frac{(x~-~\bar{X})^2}{n}$

$\implies$   $(x~-~\bar{X})^{2} ~=~ 16 ,~ 9,~ 0,~ 0,~ 4, ~25 ~respectively$

$\implies$$Variance ~=~ \frac{16 + ~9 ~+~0~+~0~+~4~+~25}{6}$

$\implies$$Variance ~=~ \frac{54}{6} ~=~ 9 .$

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Now, $Standard~ Deviation ~=~\sqrt{Variance}$

$\implies ~Standard~ Deviation ~=~\sqrt{9} ~=~ 3 .$

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$Coefficient ~ of~Variation ~=~ \frac{s}{\bar{x}} ~*~ 100$

where. $s$ = Standard Deviation of sample

and $\bar{x} ~=~$ Mean of sample .

$\implies ~Coefficient ~ of~Variation ~=~ \frac{3}{7} ~*~ 100$

$\implies ~Coefficient ~ of~Variation ~=~ 42.857$ .

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For computing Z-scores,

$Z~= ~ \frac{x ~-~ \mu}{\sigma}$

For , $Z_{1}~= ~ \frac{3 ~-~ 7}{3} ~=~\frac{-4}{3} ~= ~ -1.333$

Similarly for $Z_{2}~= ~ \frac{4 ~-~ 7}{3} ~=~\frac{-3}{3} ~= ~ -1$

$Z_{3}~= ~ \frac{7 ~-~ 7}{3} ~=~\frac{0}{3} ~= ~ 0$

$Z_{4}~= ~ \frac{7 ~-~ 7}{3} ~=~\frac{0}{3} ~= ~ 0$

$Z_{5}~= ~ \frac{9~-~ 7}{3} ~=~\frac{2}{3} ~= ~ 0.666$

$Z_{6}~= ~ \frac{12~-~ 7}{3} ~=~\frac{5}{3} ~= ~ 1.666$

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In the given data :   $3,~ 4,~ 7,~ 7,~ 9,~ 12$

$Q_{3}~= ~ 9 (i.e. ~ 3^{rd}~ Quartile)$ because it divide the right half section in to two equal sections

and $Q_{1}~= ~ 4 ~(i.e. ~ 1^{st}~ Quartile)$ which divides the left half section in to two equal sections .

Therefore, $IQR~ (Inter ~Quartile~range) ~=~ Q_{3}~-~ Q_{1}$

$\implies ~IQR~=~ 9 ~-~ 4 ~=~5$

and for being an outlier . if is greater than $Q_{3}~+ ~1.5(IQR)$

or less than $Q_{1}~- ~1.5(IQR)$ .

So, let us check out this :

$Q_{3}~+ ~1.5(IQR)$$= ~ 9 ~+~1.5(5) ~= ~~9~+~7.5~=~16.5$

$Q_{1}~- ~1.5(IQR)$$= ~ 4 ~-~1.5(5) ~= ~~4~-~7.5~=~-3.5$

That the observation which is greater than 16.5 or less than -3.5 is an outlier

but we can see that there is not any outlier .

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In the above data , $Mean ~=~Median ~=~ Mode ~=~7$

Hence , the data set is bell shaped .