Question

Need help please!! Will give points

Answer 1

Answer:

1. $\bar X=5.10$

2. $\sigma=0.47$

3.$\sigma^2=0.22$

Step-by-step explanation:

QUESTION 1

The given data set for the expenditure is

$4.85,5.10,5.50,4.75,4.50,5.00,6.00$

The formula for calculating the mean is given by,

$\bar X=\frac{\sum x}{n}$

We need to add all the expenditure and divide by the total number of days.

$\bar X=\frac{4.85+5.10+5.50+4.75+4.50+5.00+6.00}{7}$

This gives us,

$\bar X=\frac{35.7}{7}$

$\Rightarrow \bar X=5.10$ to the nearest hundredth.

QUESTION 2

The standard  deviation of the data set is given by the formula;

$\sigma =\sqrt{\frac{\sum(x-\bar X)^2}{n} }$

This implies that,

$\sigma =\sqrt{\frac{(4.85-5.10)^2+(5.10-5.10)^2+(5.50-5.10)^2+(4.75-5.10)^2+(4.50-5.10)^2+(5.00-5.10)^2+(6.00-5.10)^2}{7} }$

This will give us,

$\sigma =\sqrt{\frac{(-0.25)^2+(0.00)^2+(0.40)^2+(-0.35)^2+(-0.60)^2+(-0.10)^2+(0.90)^2}{7} }$

$\sigma =\sqrt{\frac{0.0625+0+0.16+0.1225+0.36+0.01+0.81}{7} }$

$\sigma =\sqrt{\frac{61}{280} }$

$\sigma =0.466775$

to the nearest hundredth,

$\sigma =0.47$.

QUESTION 3

The variance is the square of the standard deviation.

$\sigma^2 =(0.46675)^2$

$\sigma^2 =0.217857$

To the nearest hundred gives,

$\sigma^2 =0.22$