Need help please!! Will give points

Question

asked Aug 7, 2019 23.5k views

1 Answer

Nniloc · Answer 1 · 2019-08-13T11:48:47+0000

Answer:

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,

$\sigma^2 =0.22$