Question

The city of Madison regularly checks the quality of water at swimming beaches located on area lakes. Fifteen times the concentration of fecal coliforms, in number of colony forming units (CFU) per 100ml of water, was measured during the summer at one beach. 180 1600 90 140 50 260 400 90 380 110 10 60 20 340 80. What is the sample standard deviation?

Answer 1

Answer:

The sample standard deviation is 393.99

Step-by-step explanation:

The standard deviation of a sample can be calculated using the following formula:

$s=\sqrt[ ]{\frac{1}{N-1} \sum_{i=1}^{N}(x_{i}-{\displaystyle \textstyle {\bar {x}}}) ^{2} }$

Where:

$s=$ Sample standart deviation

$N=$ Number of observations in the sample

${\displaystyle \textstyle {\bar {x}}}=$ Mean value of the sample

and $\sum_{i=1}^{N}(x_{i}-{\displaystyle \textstyle {\bar {x}}}) ^{2} }$ simbolizes the addition of the square of the difference between each observation and the mean value of the sample.

Let's start calculating the mean value:

$\bar {x}=\frac{1}{N} \sum_{i=1}^{N}x_{i}$

$\bar {x}=\frac{1}{15}*(180+1600+90+140+50+260+400+90+380+110+10+60+20+340+80)$

$\bar {x}=\frac{1}{15}*(3810)$

$\bar {x}=254$

Now, let's calculate the summation:

$\sum_{i=1}^{N}(x_{i}-\bar {x}) ^{2} }=(180-254)^2+(1600-254)^2+(90-254)^2+...+(80-254)^2$

$\sum_{i=1}^{N}(x_{i}-\bar {x}) ^{2} }=2173160$

So, now we can calculate the standart deviation:

$s=\sqrt[ ]{\frac{1}{N-1} \sum_{i=1}^{N}(x_{i}-{\displaystyle \textstyle {\bar {x}}}) ^{2} }$

$s=\sqrt[ ]{\frac{1}{15-1}*(2173160)}$

$s=\sqrt[ ]{\frac{2173160}{14}}$

$s=393.99$

The sample standard deviation is 393.99