Question

The yield of wheat and rice per acre for 10 districts of a state is as under

$\begin{array}{ccccccccccc}\text{District}& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ \text{Wheat}& 12& 10& 15& 19& 21& 16& 18& 9& 25& 10\\ \text{Rice}& 22& 29& 12& 23& 18& 15& 12& 34& 18& 2\end{array}$

Calculate for each crop,

(i) Range

(ii) QD

(iii) Mean Deviation about Mean

(iv) Mean Deviation about Median

(v) Standard Deviation

(vi) Which crop has greater variation ?

(vii) Compare the value of different measures of each crop.

Answer 1

(i) Range

(a) Wheat: Highest value of distribution (H) =25

Lowest value of distribution (L) =9

Range=H-L=25-9=16

(b) Rice: Highest value of distribution (H) =34

Lowest value of distribution (L) =12

Range=H-L=34-12=22

(ii) Quartile Deviation

(a) Wheat: Arranging the production of wheat in increasing order

9,10,10,12,15,16,18,19,21,25

$Q_1=\frac{N+1}{4}th\text{item}=\frac{10+1}{4}th\text{item}=\frac{11}{4}th\text{item}$

=2.75th item

=Size of 2nd item+0.75 (size of 3rd item-size of 2nd item)

=10+0.75(10−10)

$=10+0.75\times 0=10$

$Q_3=\frac{3(N+1)}{4}th\text{item}=\frac{3(10+1)}{4}th\text{item}$

$=\frac{11}{4}th\text{item}$

=Size of 8th item+0.25 (size of 9th item-size of 8th item)

=19+0.25(21−19)

$=19+0.25\times 2$

=19+0.50=19.50

Quartile Deviation=$\frac{Q_3-Q_1}{2}$ = $\frac{19.50-10}{2}=\frac{9.50}{2}=4.75$

(b) Rice: Arranging the production of rice in increasing order

12,12,12,15,18,18,22,23,29,34

$Q_1=\frac{N+1}{4}th\text{item}=\frac{10+1}{4}th\text{item}$

= 2.75th item

= Size of 2nd item + 0.75 (size of 3rd item-size of 2nd item)

=12+0.75(12−12)=12+0.75$\times 0$

=12

$Q_3=\frac{3(N+1)}{4}thitem$

=$\frac{3(10+1)}{4}thitem$

=$\frac{33}{4}thitem$ = 8.25th

=8.25th item

=Size of 8th item+0.25 (size of 9th item-size of 8th item)

=23+0.25(29−23)

=23+$0.25\times 6$

=23+1.5

=24.5

Quartile Deviation= $\frac{Q_3-Q_1}{2}=\frac{24.5-12}{2}=\frac{12.50}{2}=6.25$

(iii) Mean Deviation about Mean

(a) Wheat

$\begin{array}{cc}\text{Wheat Production (X)}& |d|=|X-A\overline{X}|\\ & A\overline{X}=15\\ 9& 6\\ 10& 5\\ 10& 5\\ 12& 3\\ 15& 0\\ 16& 1\\ 18& 3\\ 19& 4\\ 21& 6\\ 25& 10\\ \sum \overline{X}=155& \sum |d|=43\end{array}$

Mean=$\frac{\sum X}{N}=\frac{155}{10}=15.5$

Mean Deviation from Mean

$MD\left(\overline{X}\right)=\frac{\sum |d|+(\overline{X}-A\overline{X})(\sum f_B-\sum f_A)}{N}$

$=\frac{43+(15.5-15)(5-5)}{10}=\frac{43}{10}=4.3$

(b) Rice

$\begin{array}{cc}\text{Wheat Production (X)}& |d|=|X-A\overline{X}|\\ & A\overline{X}=18\\ 12& 6\\ 12& 6\\ 12& 6\\ 15& 6\\ 15& 3\\ 18& 0\\ 18& 0\\ 22& 4\\ 23& 5\\ 29& 11\\ 34& 16\\ \sum \overline{X}=195& \sum |d|=57\end{array}$

(iv) Mean Deviation about Median

(a) Wheat

$\begin{array}{cc}\text{Production of wheat (X)}& |d|=|X-15|\\ 9& 6\\ 10& 5\\ 10& 5\\ 12& 3\\ 15& 0\\ 16& 1\\ 18& 3\\ 19& 4\\ 21& 6\\ 25& 10\\ & \sum |d|=43\end{array}$

Median=Size of $\left(\frac{N+1}{2}\right)$th item=Size of $\left(\frac{10+1}{2}\right)$th item

=size of (5.5)th item

=$\frac{\text{Size of 5th item}+\text{Size of 6th item}}{2}$

=$\frac{15+16}{2}=15.5$

$M{D}_{Median}=\frac{\sum |d|+(Median-A)(\sum f_b-\sum f_A)}{n}$

=$\frac{57+(18-18)(6-4)}{10}=\frac{57}{10}=5.7$

(b) Rice

$\begin{array}{cc}\text{Production of Rice (X)}& d=X-18\\ 12& 6\\ 12& 6\\ 12& 6\\ 15& 3\\ 18& 0\\ 18& 0\\ 22& 4\\ 23& 5\\ 29& 11\\ 34& 16\\ & \sum |d|=57\end{array}$

Since n is even.

Therefore, Median=size of $\left(\frac{N+1}{2}\right)$th item

=size of $\left(\frac{10+1}{2}\right)$th item

$RightarrowSizeof\left(5.5\right)thitem=\frac{\text{Size of 5th item}+\text{Size of 6th item}}{2}$

=$\frac{36+2}{2}=18$

$M{D}_{Median}=\frac{\sum |d|+(Median-A)(\sum f_b-\sum f_A)}{n}$

=$\frac{57+(18-18)(6-4)}{10}$

=$\frac{57}{10}=5.7$

(v) Standard Deviation

(a) Wheat

$\sigma =\sqrt{\frac{\sum d^2}{n}-{\left(\frac{\sum d}{n}\right)}^2}$

$=\sqrt{\frac{257}{10}-{\left(\frac{5}{10}\right)}^2}$

$=\sqrt{\frac{257}{10}-\frac{25}{100}}$

$=\sqrt{\frac{2570-25}{100}}$

5.04

(b) Rice

$\begin{array}{ccc}\text{Production of Rice (X)}& AX=18& d^2\\ & d=X-AX& \\ 12& -6& 36\\ 12& -6& 36\\ 12& -6& 36\\ 15& -3& 9\\ 18& 0& 0\\ 22& 4& 16\\ 23& 5& 25\\ 29& 11& 121\\ 34& 16& 256\\ & \sum |d|=15& \sum d^2=535\end{array}$

$\sigma =\sqrt{\frac{\sum d^2}{n}-{\left(\frac{\sum d}{n}\right)}^2}$

$=\sqrt{\frac{535}{10}-{\left(\frac{15}{10}\right)}^2}$

$=\sqrt{\frac{535}{10}-\frac{225}{100}}$

$=\sqrt{51.25}$

=7.16

(vi) Coefficient of Variation

(a) Wheat

$CV=\frac{\sigma }{\overline{X}}\times 100=\frac{5.04}{15.5}\times 100=32.51$

(b) Rice

$CV=\frac{\sigma }{\overline{X}}\times 100=\frac{7.16}{19.5}\times 100=36.71$

Rice crop has the greater variation as the coefficient of variation is higher for rice as compared to that of wheat.

(vii) Rice crop has the higher range, quartile deviation, mean deviation about mean, mean deviation about median, standard deviation and coefficient of variation.