Question

Question Number 03: According to the research conducted by the geologists of University of Texas, the temperatures of soil in \u00b0C 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 Harmonic Mean (iii) Geometric Mean (iv) Root Mean Squares

Answer 1

n = 15

i) Arithmetic Mean

$$A=\frac{1}{n}\sum x$$

$A=\frac{68.4+72.3+58.7+48.3+53.8+73.2+37.9+55.4+42.7+84.6+67.4+58.5+92.4+62.5+87.8}{15} = 64.26$

ii) Harmonic Mean

$$H=\frac{n}{\sum {\frac{1}{x}}}$$

$H=\frac{15}{\frac{1}{68.4}+\frac{1}{72.3}+\frac{1}{58.7}+\frac{1}{48.3}+\frac{1}{53.8}+\frac{1}{73.2}+\frac{1}{37.9}+\frac{1}{55.4}+\frac{1}{42.7}+\frac{1}{84.6}+\frac{1}{67.4}+\frac{1}{58.5}+\frac{1}{92.4}+\frac{1}{62.5}+\frac{1}{87.8}}=60.42$

iii) Geometric Mean

$$\bar x_g=(\prod x)^{\frac{1}{n}}$$

$\bar x_g=\sqrt[n]{68.4\cdot72.3\cdot...\cdot87.8}=62.35$

iv) Root Mean Squares

$$RMS=\sqrt{\frac{1}{n}\sum x^2}$$

$RMS=\sqrt{\frac{1}{15}(68.4^2+72.3^2+58.7^2+...+87.8^2)}=66.1$