Question #281822

Find mean, median, mean deviation about mean, mean deviation about median for the


following data


Age in Years 20 – 22 22 – 24 24 – 26 26 – 28 28 – 30 30 – 32 32 – 34


No. of Employees 70 90 110 140 130 80 80


15. Find mean deviation from mode and its coefficient for the following data


Class – Interval 14 – 18 18 – 22 22 – 26 26 – 30 30 – 34


Frequencies 5 12 14 12 7

1
Expert's answer
2022-01-02T17:42:38-0500

14.

n=700n=700


mean=xˉ=ifixin=18960700=27.0857mean=\bar{x}=\dfrac{\sum _if_ix_i}{n}=\dfrac{18960}{700}=27.0857ClassfiMid valuefixixixˉxixˉfixi2022702114706.08574262224902320704.0857367.714324261102527502.0857229.428626281402737800.08571228301302940301.9143248.85713032803126403.9143313.14293234803328005.9143473.1429700189602070.2857\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} Class & f_i & Mid\ value & f_ix_i & |x_i-\bar{x}| & |x_i-\bar{x}|f_i\\ & & x_i & & & & \\ \hline 20-22 & 70 & 21 & 1470 & 6.0857 & 426 \\ \hdashline 22-24 & 90 & 23 & 2070 & 4.0857 & 367.7143 \\ \hdashline 24-26 & 110 & 25 & 2750 & 2.0857 & 229.4286 \\ \hdashline 26-28 & 140 & 27 & 3780 & 0.0857 & 12 \\ \hdashline 28-30 & 130 & 29 & 4030 & 1.9143 & 248.8571 \\ \hdashline 30-32 & 80 & 31 & 2640 & 3.9143 & 313.1429 \\ \hdashline 32-34 & 80 & 33 & 2800 & 5.9143 & 473.1429 \\ \hdashline & 700 & & 18960 & & 2070.2857 \\ \hdashline \end{array}

Mean deviation of Mean


δxˉ=ifixixˉn=2070.2857700=2.9576\delta \bar{x}=\dfrac{\sum _if_i|x_i-\bar{x}|}{n}=\dfrac{2070.2857}{700}=2.9576

Coefficient of Mean deviation


δxˉxˉ=2.957627.0857=0.1092\dfrac{\delta \bar{x}}{\bar{x}}=\dfrac{2.9576}{27.0857}=0.1092




median=Mmedian=M

=value of (n2)th observation=\text{value of } (\dfrac{n}{2})^{th}\text{ observation}

=value of (350)th observation=\text{value of } (350)^{th}\text{ observation}

The median class is 2628.26-28.

L=L= lower boundary point of median class =26=26


n=n= Total frequency =700=700


cf=cf= Cumulative frequency of the class preceding the median class =270=270


f=f= Frequency of the median class =140=140


c=c= class length of median class =2=2


Median 

M=L+n2cffc=26+70022701402M=L+\dfrac{\dfrac{n}{2}-cf}{f}\cdot c=26+\dfrac{\dfrac{700}{2}-270}{140}\cdot 2

=27.1429=27.1429


ClassfiMid valuecfxiMxiMfixi20227021706.1429430222490231604.1429372.85712426110252702.1429235.71432628140274100.1429202830130295401.8571241.4286303280316203.8571308.5714323480337005.8571468.57147002077.1429\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} Class & f_i & Mid\ value & cf & |x_i-M| & |x_i-M|f_i\\ & & x_i & & & & \\ \hline 20-22 & 70 & 21 & 70 & 6.1429 & 430 \\ \hdashline 22-24 & 90 & 23 & 160 & 4.1429& 372.8571 \\ \hdashline 24-26 & 110 & 25 & 270 & 2.1429 & 235.7143 \\ \hdashline 26-28 & 140 & 27 & 410 & 0.1429 & 20 \\ \hdashline 28-30 & 130 & 29 & 540 & 1.8571 & 241.4286 \\ \hdashline 30-32 & 80 & 31 & 620 & 3.8571 & 308.5714 \\ \hdashline 32-34 & 80 & 33 & 700 & 5.8571& 468.5714 \\ \hdashline & 700 & & & & 2077.1429 \\ \hdashline \end{array}

Mean deviation of Median


δxˉ=ifixiMn=2077.1429700=2.9673\delta \bar{x}=\dfrac{\sum _if_i|x_i-M|}{n}=\dfrac{2077.1429}{700}=2.9673

Coefficient of Mean deviation


δxˉxˉ=2.967327.1429=0.1093\dfrac{\delta \bar{x}}{\bar{x}}=\dfrac{2.9673}{27.1429}=0.1093



15.


ClassfiMid valuexiZxiZfixi141851684018221220448222614240026301228448303473285650192\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} Class & f_i & Mid\ value & |x_i-Z| & |x_i-Z|f_i\\ & & x_i & & & \\ \hline 14-18 & 5 & 16 & 8 & 40 \\ \hdashline 18-22 & 12 & 20 & 4 & 48 \\ \hdashline 22-26 & 14 & 24 & 0 & 0 \\ \hdashline 26-30 & 12 & 28 & 4 & 48 \\ \hdashline 30-34 & 7 & 32 & 8 & 56 \\ \hdashline & 50 & & & 192 \\ \hdashline \end{array}

To find Mode Class

Here, maximum frequency is 14.14.


The mode class is 2226.22-26.


L=L= lower boundary point of mode class =22=22


f1=f_1=  frequency of the mode class =14=14


f0=f_0=  frequency of the preceding class =12=12


f2=f_2=  frequency of the succedding class =12=12


c=c=  class length of mode class =4=4


Z=L+(f1f02f1f0f2)cZ=L+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})\cdot c

=22+(14122(14)1212)4=24=22+(\dfrac{14-12}{2(14)-12-12})\cdot 4=24

Mean deviation of Mode


δxˉ=ifixiZn=19250=3.84\delta \bar{x}=\dfrac{\sum _if_i|x_i-Z|}{n}=\dfrac{192}{50}=3.84

Coefficient of Mean deviation


δxˉxˉ=3.8424=0.16\dfrac{\delta \bar{x}}{\bar{x}}=\dfrac{3.84}{24}=0.16




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