Answer to Question #89613 in Statistics and Probability for Sumendra

Question

Table 1
3.80 3.77 3.70 3.74 3.70
3.86 3.76 3.68 3.67 3.57
3.83 3.70 3.80 3.74 3.67
3.78 3.74 3.73 3.65 3.66
3.75 3.64 3.78 3.73 3.64

Based on Table 1,

a). Calculate the variance
b). Calculate the interquartile range

Answer 1

$$3.57,\ 3.64,\ 3.64,\ 3.65,\ 3.66,$$ $$3.67,\ 3.67,\ 3.68,\ 3.70,\ 3.70,$$ $$3.70,\ 3.73,\ 3.73,\ 3.74,\ 3.74,$$ $$3.74,\ 3.75,\ 3.76,\ 3.77,\ 3.78,$$ $$3.78,\ 3.80,\ 3.80,\ 3.83,\ 3.86.$$

$$Var(X)=\sigma^2={1 \over N}\displaystyle\sum_{i=1}^N(x_i-\overline{x})^2$$

$$\overline{x}={1 \over N}\displaystyle\sum_{i=1}^Nx_i$$ $$\overline{x}=3.7236$$

$$\displaystyle\sum_{i=1}^N(x_i-\overline{x})^2=0.109376$$ $$Var(X)=\sigma^2=0.00437504$$

The inter-quartile range $IQR$ is the distance from the third quartile (75th percentile) to first quartile (25th percentile) on a frequency distribution. 

The first quartile $Q_1$ is the point which gives us 25% of the area to the left of it and 75% to the right of it.

The third quartile $Q_3$ is the point which gives us 75% of the area to the left of it and 25% of the area to the right of it. 

$$IQR=Q_3-Q_1$$

$$Q_1={3.67+3.67 \over 2}=3.67$$

$$Q_3={3.77+3.78 \over 2}=3.775$$

$$IQR=3.775-3.67=0.105$$