Answer to Question #109866 in Statistics and Probability for julaine

Question

For a random sample of 12 married couples the age difference (male age minus female age) in years is recorded below. For the below data calculate;
3 0 -4 7 12 2
4 -2 1 20 1 6

For this data calculate the ;
mean
range
interquartile
mean deviation

show all working

Answer 1

Lets arrange of the values in an ascending order:

$-4,-2,0,1,1,2,3,4,6,7,12,20$

$$mean=\bar{x}={1 \over N}\displaystyle\sum_{i=1}^Nx_i=$$

$$={-4-2+0+1+1+2+3+4+6+7+12+20 \over 12}={25 \over 6}\approx4.666667$$

$$Range=20-(-4)=24$$

1st Quartile (Q₁): $\dfrac{0+1}{2}=0.5$

2nd Quartile (Q₂): $\dfrac{2+3}{2}=2.5$

3rd Quartile (Q₃): $\dfrac{6+7}{2}=6.5$

$IQR = Q_3 - Q_1 = 6.5 - 0.5 = 6$

Or

$Q_1=(\dfrac{12+1}{4})^{th}Term=0$

$Q_3=(\dfrac{3(12+1)}{4})^{th}Term=7$

$IQR = Q_3 - Q_1 = 7-0 = 7$

$$\begin{matrix} x_i & x_i-\bar{x} & |x_i-\bar{x}| \\ -4 & -49/6 & 49/6 \\ -2 & -37/6 & 37/6\\ 0 & -25/6 & 25/6\\ 1 & -19/6 & 19/6\\ 1 & -19/6 & 19/6\\ 2 & -13/6 & 13/6\\ 3 & -7/6 & 7/6\\ 4 & -1/6 & 1/6\\ 6 & 11/6 & 11/6\\ 7 & 17/6 & 17/6\\ 12 & 47/6 & 47/6\\ 20 & 95/6 & 95/6 \end{matrix}$$

$$mean\ deviation={1 \over N}\displaystyle\sum_{i=1}^N|x_i-\bar{x}|=$$

$$={1 \over 12}({340 \over 6})={85 \over 18}\approx4.722222$$