Answer to Question #224165 in Statistics and Probability for Khan

Question

Question Number 03: [05 Marks]

According to the research conducted by the geologists of University of Texas, the temperatures of soil in 0C at 750 meters deep inside the ground on different locations of earth are mentioned below:

68.4 72.3 58.7 48.3 53.8 73.2 37.9 55.4 42.7 84.6 67.4 58.5 92.4 62.5 87.8

Calculate:

(i) Arithmetic Mean

(ii) Median

(iii) Mode

(iv) Variance

(v) Standard Deviation

Answer 1

$solution\\ given \space data \space set \\ 68.4, 72.3 ,58.7 ,48.3 ,53.8 ,73.2 ,37.9 ,55.4 ,42.7 ,84.6 ,67.4 ,58.5 ,92.4 ,62.5 ,87.8\\ ---------------------------------\\ (i) Arithmetic Mean\\ mean=\frac{\sum x_i}{n}=\frac{68.4+ 72.3 +58.7 +48.3 +53.8 +73.2 +37.9 +55.4 +42.7 +84.6 +67.4 +58.5 +92.4 +62.5 +87.8}{15}=\frac{963.9}{15}=64.26\\ \\ ---------------------------------\\(ii) Median\\ rearrange \space data \space set \space as \space ascending \space order \\ 37.9, 42.7, 48.3, 53.8, 55.4, 58.5, 58.7, 62.5, 67.4, 68.4, 72.3, 73.2, 84.6, 87.8, 92.4\\ now \space mid \space term \space value \space is \space median \space =62.5\\ ---------------------------------\\(iii) Mode\\All \space values \space appeared \space just \space once\\ hence \space there \space is \space no \space mode\\ ---------------------------------\\(iv) Variance(s^2)\\ s^{2} = \dfrac{\sum_{i=1}^{n}(x_i - \overline{x})^{2}}{n - 1}\\ where \space x_i \space is \space data \space \space values \space and \space \space \bar{x } \space is \space mean \\ s^{2} = \dfrac{(68.4 -64.26)^{2}+(72.3 -64.26)^{2}+...+(58.7-64.26)^{2}}{15 - 1}\\ s^2=257.19829\\ ---------------------------------\\(v) Standard Deviation\\ standard \space deviation \space is \space the \space root \space of \space variance\\ standard \space deviation = \sqrt{variance}\\ standard \space deviation = \sqrt{257.19829}\\ standard \space deviation = 16.037\\$