Question

Need help please!! Will give points

Answer 1

1.
`\bar X=5.10`

2.
`\sigma=0.47`

3.
`\sigma^2=0.22`

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=(\sum x)/(n)`

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

`\bar X=(4.85+5.10+5.50+4.75+4.50+5.00+6.00)/(7)`

This gives us,

`\bar X=(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{(\sum(x-\bar X)^2)/(n) }`

This implies that,

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

`\sigma =\sqrt{(0.0625+0+0.16+0.1225+0.36+0.01+0.81)/(7) }`

`\sigma =\sqrt{(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`