Question #281821

Find the mean deviation from mean and median for the following data


13, 17, 9, 21, 15, 25, 30, 35, 31


12. Find the mean deviation from mean and its coefficient for the following data representing


age distribution of 50 boys.


Age (in years): 17 18 19 20 21 22 23


No. of Boys 8 12 15 8 4 2 1


13. Find the mean deviation from median and its coefficient for the following data:


Height (in cm) 160 164 168 172 176 180


Number of students 3 7 12 15 6 2

1
Expert's answer
2021-12-31T08:31:15-0500

11.


9,13,15,17,21,25,30,31,359,13, 15,17, 21, 25, 30, 31, 35

mean=xˉ=ixin=19(13+17+9+21mean=\bar{x}=\dfrac{\sum _ix_i}{n}=\dfrac{1}{9}(13+17+9+21

+15+25+30+35+31)=196921.7778+15+25+ 30+35+31)=\dfrac{196}{9}\approx21.7778

median=M=21median=M=21

xixixˉxiM912.777812138.77788156.77786174.77784210.77780253.22224308.22229319.2222103513.222214sum=19667.777867\def\arraystretch{1.5} \begin{array}{c:c:c:c} & x_i & |x_i-\bar{x}| & |x_i-M| \\ \hline & 9 & 12.7778 &12 \\ \hdashline & 13 & 8.7778 & 8 \\ \hdashline & 15 & 6.7778 & 6 \\ \hdashline & 17 & 4.7778 & 4 \\ \hdashline & 21 & 0.7778 & 0 \\ \hdashline & 25 & 3.2222 & 4 \\ \hdashline &30 & 8.2222 & 9 \\ \hdashline & 31 & 9.2222 & 10 \\ \hdashline & 35 & 13.2222 &14 \\ \hdashline sum= & 196 & 67.7778 & 67 \\ \hdashline \end{array}

Mean deviation of Mean


δxˉ=ixixˉn=67.77789=7.5309\delta \bar{x}=\dfrac{\sum _i|x_i-\bar{x}|}{n}=\dfrac{67.7778}{9}=7.5309

Mean deviation of Median


δxˉ=ixiMn=679=7.4444\delta \bar{x}=\dfrac{\sum _i|x_i-M|}{n}=\dfrac{67}{9}=7.4444

12.


mean=xˉ=ifixin=94850=18.96mean=\bar{x}=\dfrac{\sum _if_ix_i}{n}=\dfrac{948}{50}=18.96xififixixixˉxixˉfi1781361.9615.6818122160.9611.5219152750.040.62081601.048.32214842.048.16222443.046.08231234.044.045094854.4\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & x_i & f_i & f_ix_i & |x_i-\bar{x}| & |x_i-\bar{x}|\cdot f_i \\ \hline & 17 & 8 &136 & 1.96 & 15.68 \\ \hdashline & 18 & 12 & 216 & 0.96 & 11.52 \\ \hdashline & 19 & 15 & 275 & 0.04 & 0.6 \\ \hdashline & 20 & 8 & 160 & 1.04 & 8.32 \\ \hdashline & 21 & 4 & 84 & 2.04 & 8.16 \\ \hdashline & 22 & 2 & 44 & 3.04 & 6.08 \\ \hdashline & 23 & 1 & 23 & 4.04 & 4.04 \\ \hdashline & & 50 & 948 & & 54.4 \\ \hdashline \end{array}

Mean deviation of Mean


δxˉ=ifixixˉn=54.450=1.088\delta \bar{x}=\dfrac{\sum _if_i|x_i-\bar{x}|}{n}=\dfrac{54.4}{50}=1.088

Coefficient of Mean deviation


δxˉxˉ=1.08818.96=0.0574\dfrac{\delta \bar{x}}{\bar{x}}=\dfrac{1.088}{18.96}=0.0574

13.


median=Mmedian=M

=value of (n+12)th observation=\text{value of } (\dfrac{n+1}{2})^{th}\text{ observation}

=value of (23)th observation=172=\text{value of } (23)^{th}\text{ observation}=172


xificfxiMxiMfi160331236164710856168122244817215370017664342418024581645180\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & x_i & f_i & cf & |x_i-M| & |x_i-M|\cdot f_i \\ \hline & 160 & 3 & 3 & 12 & 36 \\ \hdashline & 164 & 7 & 10 & 8 & 56 \\ \hdashline & 168 & 12 & 22 & 4 & 48 \\ \hdashline & 172 & 15 & 37 & 0 & 0 \\ \hdashline & 176 & 6 & 43 & 4 & 24 \\ \hdashline & 180 & 2 & 45 & 8 & 16 \\ \hdashline & & 45 & & & 180 \\ \hdashline \end{array}

Mean deviation of Median


δxˉ=ifixiMn=18045=4\delta \bar{x}=\dfrac{\sum _if_i|x_i-M|}{n}=\dfrac{180}{45}=4

Coefficient of Mean deviation


δxˉxˉ=4172=0.0233\dfrac{\delta \bar{x}}{\bar{x}}=\dfrac{4}{172}=0.0233




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS