Statistical file characteristics

Determine the median, mode, arithmetic mean, variance, standard deviation, range of variation, and coefficient of variation of the character x in the statistical file:
2x 9, 7x 10, 9x 11, 11x 15, 15x 17, 16x 19, 13x 21, 10x 25, 9x 29, 4x 32

Final Answer:

m₁ =  19
m₂ =  19
μ =  18.9896
σ =  6.0956
r =  23
k =  0.321

Step-by-step explanation:

$$\begin{gathered} 2\cdot 9,7\cdot 10,9\cdot 11,11\cdot 15,15\cdot 17,16\cdot 19,13\cdot 21,10\cdot 25,9\cdot 29,4\cdot 32 \\ \{9,9, \\ 10,10,10,10,10,10,10, \\ 11,11,11,11,11,11,11,11,11, \\ 15,15,15,15,15,15,15,15,15,15,15, \\ 17,17,17,17,17,17,17,17,17,17,17,17,17,17,17, \\ 19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19, \\ 21,21,21,21,21,21,21,21,21,21,21,21,21, \\ 25,25,25,25,25,25,25,25,25,25, \\ 29,29,29,29,29,29,29,29,29, \\ 32,32,32,32\} \\ {m}_1=19 \end{gathered}$$

$$\begin{gathered} m_2=modus \\ {m}_2=19 \end{gathered}$$

$$\begin{gathered} n=2+7+9+11+15+16+13+10+9+4=96 \\ s=2\cdot 9+7\cdot 10+9\cdot 11+11\cdot 15+15\cdot 17+16\cdot 19+13\cdot 21+10\cdot 25+9\cdot 29+4\cdot 32=1823 \\ \mu =\frac{s}{n}=\frac{1823}{96}=18\frac{95}{96}=18.9896 \end{gathered}$$

$$\begin{gathered} s_2=2\cdot (9-\mu {)}^2+7\cdot (10-\mu {)}^2+9\cdot (11-\mu {)}^2+11\cdot (15-\mu {)}^2+15\cdot (17-\mu {)}^2+16\cdot (19-\mu {)}^2+13\cdot (21-\mu {)}^2+10\cdot (25-\mu {)}^2+9\cdot (29-\mu {)}^2+4\cdot (32-\mu {)}^2=2\cdot (9-18.9896{)}^2+7\cdot (10-18.9896{)}^2+9\cdot (11-18.9896{)}^2+11\cdot (15-18.9896{)}^2+15\cdot (17-18.9896{)}^2+16\cdot (19-18.9896{)}^2+13\cdot (21-18.9896{)}^2+10\cdot (25-18.9896{)}^2+9\cdot (29-18.9896{)}^2+4\cdot (32-18.9896{)}^2\doteq 3566.9896 \\ \sigma =\sqrt{\frac{s_2}{n}}=\sqrt{\frac{3566.9896}{96}}=6.0956 \end{gathered}$$

$$\begin{gathered} r=\max -min \\ r=32-9=23 \end{gathered}$$

$k=\frac{\sigma }{\mu }=\frac{6.0956}{18.9896}=0.321$

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