Answer:
1. [tex]\bar X=5.10[/tex]
2. [tex]\sigma=0.47[/tex]
3.[tex]\sigma^2=0.22[/tex]
Step-by-step explanation:
QUESTION 1
The given data set for the expenditure is
[tex]4.85,5.10,5.50,4.75,4.50,5.00,6.00[/tex]
The formula for calculating the mean is given by,
[tex]\bar X=\frac{\sum x}{n}[/tex]
We need to add all the expenditure and divide by the total number of days.
[tex]\bar X=\frac{4.85+5.10+5.50+4.75+4.50+5.00+6.00}{7}[/tex]
This gives us,
[tex]\bar X=\frac{35.7}{7}[/tex]
[tex]\Rightarrow \bar X=5.10[/tex] to the nearest hundredth.
QUESTION 2
The standard deviation of the data set is given by the formula;
[tex]\sigma =\sqrt{\frac{\sum(x-\bar X)^2}{n} }[/tex]
This implies that,
[tex]\sigma =\sqrt{\frac{(4.85-5.10)^2+(5.10-5.10)^2+(5.50-5.10)^2+(4.75-5.10)^2+(4.50-5.10)^2+(5.00-5.10)^2+(6.00-5.10)^2}{7} }[/tex]
This will give us,
[tex]\sigma =\sqrt{\frac{(-0.25)^2+(0.00)^2+(0.40)^2+(-0.35)^2+(-0.60)^2+(-0.10)^2+(0.90)^2}{7} }[/tex]
[tex]\sigma =\sqrt{\frac{0.0625+0+0.16+0.1225+0.36+0.01+0.81}{7} }[/tex]
[tex]\sigma =\sqrt{\frac{61}{280} }[/tex]
[tex]\sigma =0.466775[/tex]
to the nearest hundredth,
[tex]\sigma =0.47[/tex].
QUESTION 3
The variance is the square of the standard deviation.
[tex]\sigma^2 =(0.46675)^2[/tex]
[tex]\sigma^2 =0.217857[/tex]
To the nearest hundred gives,
[tex]\sigma^2 =0.22[/tex]
