Hello there. To solve this question, we'll have to remember some properties about mean and standard deviation.
Given the table with the calories of each pizza type.
We have to determine why the standard deviation is 87.6 and understand what does it mean in context.
First, the mean of the values were calculated in question 8, that is the arithmetic mean of the values.
Adding all calories and dividing it by the number of values, we get
[tex]\dfrac{168+181+165+177+321+380+309+313+157}{9}=\frac{2171}{9}\approx241.2[/tex]
Okay. Now, we find the variance of the set by using the formula:
[tex]Var(x)=\sum_{i=1}^n\dfrac{(x_i-\mu)^2}{n}[/tex]
In which we get
[tex]Var(x)=\dfrac{(168-241.2)^2+(181-241.2)^2+\cdots+(157-241.2)^2}{9}=7673.2[/tex]
And the standard deviation is the square root of the variance, hence
[tex]\sigma=\sqrt{Var(x)}=\sqrt{7673.2}\approx87.6[/tex]
In statistics, the standard deviation is a measure of how are the values getting away from the meanO, that is the point in the middle of the distribution:
This is the answer we're looking for.
For number eleven, suppose the new pizza has x calories.
We'll have a change in the mean, now being
[tex]\mu=\dfrac{2171+x}{10}[/tex]
And the variance will be
[tex]Var(x)=\sum_{i=1}^n\dfrac{(x_i-\mu)^2}{10}=\dfrac{\left(168-\dfrac{2171+x}{10}\right)^2+\left(181-\dfrac{2171+x}{10}\right)^2+\left(177-\dfrac{2171+x}{10}\right)^2+\left(157-\dfrac{2171+x}{10}\right)^2+\left(321-\dfrac{2171+x}{10}\right)^2+\left(309-\dfrac{2171+x}{10}\right)^2+\left(380-\dfrac{2171+x}{10}\right)^2+\left(313-\dfrac{2171+x}{10}\right)^2+\left(165-\dfrac{2171+x}{10}\right)^2+\left(x-\dfrac{2171+x}{10}\right)^2}{10}[/tex]
Consider taking x to be closer to the mean, that is
[tex]x\approx241.2[/tex]
So the mean would be
[tex]\dfrac{2171+241.2}{10}\approx241.2[/tex]
Won't change at all
And the difference between the value would be identically zero
Hence the variance will become smaller, making the standard deviation (the square root of it), smaller as well = 82.58
