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klasskru [66]
3 years ago
15

Need help please!! Will give points

Mathematics
1 answer:
Nataly [62]3 years ago
3 0

Answer:

1. \bar X=5.10

2. \sigma=0.47

3.\sigma^2=0.22


Step-by-step explanation:

<u>QUESTION 1</u>

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.


<u>QUESTION 2</u>

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.


<u>QUESTION 3</u>

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











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