Question #281818

Find the quartile deviation for the following set of 11 bottles of 1 litre of emery milk for


11 days by a dealer.


34, 37, 32, 25, 40, 45, 50, 42, 39, 47, 55


7. Calculate the quartile deviation and its coefficient from the following:


x 43 44 45 46 47 48 49


f 10 18 21 18 15 10 8

1
Expert's answer
2021-12-30T06:21:09-0500

6.


xfcf2511321234133714391540164217451847195011055111\def\arraystretch{1.5} \begin{array}{c:c:c} x & f & cf \\ \hline 25 & 1 & 1 \\ \hdashline 32 & 1 & 2 \\ \hdashline 34 & 1 & 3 \\ \hdashline 37 & 1 & 4 \\ \hdashline 39 & 1 & 5 \\ \hdashline 40 & 1 & 6 \\ \hdashline 42 & 1 & 7 \\ \hdashline 45 & 1 & 8 \\ \hdashline 47 & 1 & 9 \\ \hdashline 50 & 1 & 10 \\ \hdashline 55 & 1 & 11 \\ \hdashline \end{array}

Here, n=11.n=11.

Q1=(n+14)Q_1=(\dfrac{n+1}{4}) th value of the observation


=(11+14)=(\dfrac{11+1}{4}) th value of the observation


=(3)=(3) th value of the observation


=34=34


Q3=(3(n+1)4)Q_3=(\dfrac{3(n+1)}{4}) th value of the observation


=(3(11+1)4)=(\dfrac{3(11+1)}{4}) th value of the observation


=(9)=(9) th value of the observation


=47=47

Quartile deviation =Q3Q12=47342=6.5=\dfrac{Q_3-Q_1}{2}=\dfrac{47-34}{2}=6.5


7.


xfcf431010441828452149461867471582481092498100\def\arraystretch{1.5} \begin{array}{c:c:c} x & f & cf \\ \hline 43 & 10 & 10 \\ \hdashline 44 & 18 & 28 \\ \hdashline 45 & 21 & 49 \\ \hdashline 46 & 18 & 67 \\ \hdashline 47 & 15 & 82 \\ \hdashline 48 & 10 & 92 \\ \hdashline 49 & 8 & 100 \\ \hdashline \end{array}

Here, n=100.n=100.

Q1=(n+14)Q_1=(\dfrac{n+1}{4}) th value of the observation


=(100+14)=(\dfrac{100+1}{4}) th value of the observation


=(25.25)=(25.25) th value of the observation


=44=44


Q3=(3(n+1)4)Q_3=(\dfrac{3(n+1)}{4}) th value of the observation


=(3(100+1)4)=(\dfrac{3(100+1)}{4}) th value of the observation


=(75.75)=(75.75) th value of the observation


=47=47

Quartile deviation =Q3Q12=47442=1.5=\dfrac{Q_3-Q_1}{2}=\dfrac{47-44}{2}=1.5


Coefficient of Quartile deviation =Q3Q1Q3+Q1=474447+44=3910.033=\dfrac{Q_3-Q_1}{Q_3+Q_1}=\dfrac{47-44}{47+44}=\dfrac{3}{91}\approx0.033




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