Question #109866
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

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1
Expert's answer
2020-04-15T17:56:03-0400

Lets arrange of the values in an ascending order:

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


mean=xˉ=1Ni=1Nxi=mean=\bar{x}={1 \over N}\displaystyle\sum_{i=1}^Nx_i=

=42+0+1+1+2+3+4+6+7+12+2012=2564.666667={-4-2+0+1+1+2+3+4+6+7+12+20 \over 12}={25 \over 6}\approx4.666667

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

1st Quartile (Q1): 0+12=0.5\dfrac{0+1}{2}=0.5

2nd Quartile (Q2): 2+32=2.5\dfrac{2+3}{2}=2.5

3rd Quartile (Q3): 6+72=6.5\dfrac{6+7}{2}=6.5

IQR=Q3Q1=6.50.5=6IQR = Q_3 - Q_1 = 6.5 - 0.5 = 6


Or

Q1=(12+14)thTerm=0Q_1=(\dfrac{12+1}{4})^{th}Term=0

Q3=(3(12+1)4)thTerm=7Q_3=(\dfrac{3(12+1)}{4})^{th}Term=7


IQR=Q3Q1=70=7IQR = Q_3 - Q_1 = 7-0 = 7


xixixˉxixˉ449/649/6237/637/6025/625/6119/619/6119/619/6213/613/637/67/641/61/6611/611/6717/617/61247/647/62095/695/6\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=1Ni=1Nxixˉ=mean\ deviation={1 \over N}\displaystyle\sum_{i=1}^N|x_i-\bar{x}|=

=112(3406)=85184.722222={1 \over 12}({340 \over 6})={85 \over 18}\approx4.722222



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