Question

The consumption of number of guava and orange on a particular week by a family are given below.
Number of Guavas 3 5 6 4 3 5 4
Number of Oranges 1 3 7 9 2 6 2
Which fruit is consistently consumed by the family?

Answer 1

Answer:

Guavasare conisumed consisteoil by family.

Step-by-step explanation:

The standard deviation is a metric that reveals how much variance from the mean there is, including spread, dispersion, and spread. A "typical" variation from the mean is shown by the standard deviation. Because it uses the data set's original units of measurement, it is a well-liked measure of variability.

What one standard deviation (SD) means. on a normal or bell-shaped distribution of the data. The region of a bell curve starting at position 13 on the y axis is generally filled by 1 SD = 1 Standard deviation = 68% of the scores or data values.

We have to find co-effieat of variation for two eructs so that, the coreficient of variation (cv) is less that fruit consumed consistently

$\therefore c \cdot V=\frac{\text { STANIDARD DEVIAIION }(\sigma)}{\text { MEAN }(\bar{x})} \times 100$

For Guavas:

$\begin{array}{lcc}\text { s.No } & x_i & x_i^2 \\ \text { i) } & 3 & 9 \\ \text { 2) } & 5 & 25 \\ \text { 3) } & 6 & 36 \\ \text { 4) } & 4 & 16 \\ \text { b) } 3 & 9 \\ \text { 6) } & 5 & 25 \\ \text { Ex } & \underline{x_i=26} & \sum x_i^2=120\end{array}$

NUMBER OF GUUAS,n=6

STANDARD DEVTATIOON OF GUUAS

$\begin{aligned}\sigma_1 & =\sqrt{\frac{\varepsilon x_i^2}{n}-\left(\frac{\varepsilon x_i}{n}\right)^2} \\& =\sqrt{\frac{120}{6}-\left(\frac{26}{6}\right)^2} \\\sigma_1 & =\sqrt{20-(4.34)^2} \\& =\sqrt{1.23}\end{aligned}$

$\begin{aligned}& \dot{\sigma}=1.105 \\& {MEAN}(\bar{x})=\frac{\leq x_i}{\eta}=\frac{26}{6}=4.34 \\& \therefore C v_1=\frac{\sigma_1}{\bar{x}_1} \times 100=\frac{1.105}{4.34} \times 100 \\& C v_1=25.51 \%\end{aligned}$

FOR OR ANGES:

$\begin{aligned} \sigma_2 & =\sqrt{\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2} \\ & =\sqrt{\frac{180}{6}-\left(\frac{28}{6}\right)^2} \\ \sigma_2 & =2.867 \\ \bar{x}_2 & =\frac{\sum x_i}{n}=\frac{28}{6}=4.67\end{aligned}$

$\begin{aligned} & c v_2=\frac{\sigma_2}{\bar{x}_2}=\frac{2.867}{4.67} \\ & c v_2=61.445 \% \\ & c v_1(25.5 \%) < c v_2(61.445 \%)\end{aligned}$

Guvas are conisumed consisteoil by family.

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