# Question #ef40b

Oct 24, 2016

#### Explanation:

a) I'll do this quick since you seem to know it,

Mean
$= \frac{34550 + 45500 + 75000 + 33000 + 89000 + 40000 + 55000 + 62500 + 88000 + 81750}{10}$

Mean $= 60430$

$\sigma = \sqrt{\frac{{\left(- 25880\right)}^{2} + {\left(- 14930\right)}^{2} + {14570}^{2} + {\left(- 27430\right)}^{2} + {28570}^{2} + {\left(- 20430\right)}^{2} + {\left(- 5430\right)}^{2} + {2070}^{2} + {27570}^{2} + {21320}^{2}}{10 - 1}}$

$\sigma = 21958.081$

b)
We can assume that as we get more samples our data will approximate a normal curve with a mean of 60430 and a $\sigma$ of 21958.081.

In a normal distribution, 99.7% of samples fall within 3 standard deviations of the mean. with numbers outside this usually being considered outliers.

This would suggest that the highest value possible would be,
Mean$+ 3 \cdot \sigma$

$= 60430 + 3 \cdot 21958.081$

$= 126304.243$.

This would be the highest price based on the sample.

You mentioned 2 standard deviations, so if that is what the question suggests is the highest than that would be,

$60430 + 2 \cdot 21958.081$

$= 104346.162$ is the price two standard deviations out from the mean .

Hope this helped even though it is a bit late.