A sample 2

A sample of underweight babies was fed a special diet, and the following weight gains (lbs) were observed at the end of three months: 6.7, 2.7, 2.5, 3.6, 3.4, 4.1, 4.8, 5.9, and 8.3. Find the variance. (Round off to the nearest hundredths)

Correct answer:

v =  3.8125

Step-by-step explanation:

$$\begin{gathered} 6.7,2.7,2.5,3.6,3.4,4.1,4.8,5.9,8.3 \\ n=9 \\ s=6.7+2.7+2.5+3.6+3.4+4.1+4.8+5.9+8.3=42 \\ \mu =\frac{s}{n}=\frac{42}{9}=\frac{14}{3}=4\frac{2}{3}\doteq 4.6667 \\ {s}_2=(6.7-\mu {)}^2+(2.7-\mu {)}^2+(2.5-\mu {)}^2+(3.6-\mu {)}^2+(3.4-\mu {)}^2+(4.1-\mu {)}^2+(4.8-\mu {)}^2+(5.9-\mu {)}^2+(8.3-\mu {)}^2=(6.7-\frac{14}{3}{)}^2+(2.7-\frac{14}{3}{)}^2+(2.5-\frac{14}{3}{)}^2+(3.6-\frac{14}{3}{)}^2+(3.4-\frac{14}{3}{)}^2+(4.1-\frac{14}{3}{)}^2+(4.8-\frac{14}{3}{)}^2+(5.9-\frac{14}{3}{)}^2+(8.3-\frac{14}{3}{)}^2=\frac{61}{2}=30\frac{1}{2}=30.5 \\ v=\frac{s_2}{n-1}=\frac{\frac{61}{2}}{9-1}=\frac{61}{16}=3\frac{13}{16}=3.8125 \end{gathered}$$

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