You have found the following ages (in years) of all `plural( DATA_POINTS, animal( 1 ) )` at your local zoo:

`\qquad`

`DATA.join( ",\\enspace " )`

What is the average age of the `plural( animal( 1 ) )` at your zoo? What is the standard deviation?
You may round your answers to the nearest tenth.

Average age:`\quad`

`mean( DATA )` years old

Standard deviation:`\quad`

`stdDevPop( DATA )` years

decimals, like

`7.5`

answers within

`\pm 0.15`

are accepted to allow for rounding part-way through
Because we have data for all `plural( DATA_POINTS, animal( 1 ) )` at the zoo, we are able
to calculate the population mean
`(\color{`

and
population standard deviation `BLUE`}{\mu})`(\color{`

.
`PINK`}{\sigma})

To find the population mean, add up the values of all

ages and divide by `DATA_POINTS`

.
`DATA_POINTS`

```
\color{
```

`BLUE`}{\mu} \quad = \quad
\dfrac{\sum\limits_{i=1}^{\color{`GREEN`}{N}} x_i}{\color{`GREEN`}{N}} \quad = \quad
\dfrac{\sum\limits_{i=1}^{\color{`GREEN`}{`DATA_POINTS`}} x_i}{\color{`GREEN`}{`DATA_POINTS`}}

```
\color{
```

`BLUE`}{\mu} \quad = \quad
\dfrac{`plus.apply( KhanUtil, DATA )`}{\color{`GREEN`}{`DATA_POINTS`}} \quad = \quad
\color{`BLUE`}{`MEAN`\text{ `plural( "year", MEAN )` old}}

Find the squared deviations from the mean for each `animal(1)`.

Age

`x_i`

Distance from the mean
`(x_i - \color{``BLUE`}{\mu})

`(x_i - \color{``BLUE`}{\mu})^2

`POINT`

`roundTo( 2, POINT - MEAN )`

`SQR_DEV[ i ]`

`^2`

Because we used the population mean`(\color{`

to compute the
squared deviations from the mean, we can find the variance
`BLUE`}{\mu})`(\color{red}{\sigma^2})`

, without introducing any bias, by simply averaging the
squared deviations from the mean:

```
\color{red}{\sigma^2} \quad = \quad
\dfrac{\sum\limits_{i=1}^{\color{
```

`GREEN`}{N}} (x_i - \color{`BLUE`}{\mu})^2}{\color{`GREEN`}{N}}

```
\color{red}{\sigma^2} \quad = \quad
\dfrac{
```

`plus.apply( KhanUtil, $.map( SQR_DEV, function( x ) { return "\\color{purple}{" + x + "}"; }) )`}
{\color{`GREEN`}{`DATA_POINTS`}}

```
\color{red}{\sigma^2} \quad = \quad
\dfrac{\color{purple}{
```

`roundTo( 2, sum( SQR_DEV ) )`}}{\color{`GREEN`}{`DATA_POINTS`}} \quad = \quad
\color{red}{`VARIANCE_POP`\text{ `plural( "year", VARIANCE_POP )`}^2}

As you might guess from the notation, the population standard deviation
`(\color{`

is found by taking the square root of the population variance
`PINK`}{\sigma})`(\color{red}{\sigma^2})`

.

`\color{`

`PINK`}{\sigma} = \sqrt{\color{red}{\sigma^2}}

```
\color{
```

`PINK`}{\sigma} = \sqrt{\color{red}{`VARIANCE_POP`\text{ `plural( "year", VARIANCE_POP )`}^2}} =
\color{`PINK`}{`STDDEV_POP`\text{ `plural( "year", STDDEV_POP )`}}

**
The average animal( 1 ) at the zoo is plural( MEAN, "year" ) old with a standard deviation
of plural( STDDEV_POP, "year" ).
**

You have found the following ages (in years) of `plural( DATA_POINTS, animal( 1 ) )`
randomly selected from the `plural( POPULATION, animal( 1 ) )` at your local zoo:

`\qquad`

`DATA.join( ",\\enspace " )`

Based on your sample, what is the average age of the `plural( animal( 1 ) )`? What is the standard deviation?
You may round your answers to the nearest tenth.

Average age:`\quad`

`mean( DATA )` years old

Standard deviation:`\quad`

`stdDev( DATA )` years

decimals, like

`0.75`

answers within

`\pm 0.15`

are accepted to allow for rounding part-way through
Because we only have data for a small sample of the `plural( POPULATION, animal( 1 ) )`, we are only able
to estimate the population mean and standard deviation by finding the sample mean
`(\color{`

and
sample standard deviation `BLUE`}{\overline{x}})`(\color{`

.
`PINK`}{s})

To find the sample mean, add up the values of all

samples and divide by `DATA_POINTS`

.
`DATA_POINTS`

```
\color{
```

`BLUE`}{\overline{x}} \quad = \quad
\dfrac{\sum\limits_{i=1}^{\color{`GREEN`}{n}} x_i}{\color{`GREEN`}{n}} \quad = \quad
\dfrac{\sum\limits_{i=1}^{\color{`GREEN`}{`DATA_POINTS`}} x_i}{\color{`GREEN`}{`DATA_POINTS`}}

```
\color{
```

`BLUE`}{\overline{x}} \quad = \quad
\dfrac{`plus.apply( KhanUtil, DATA )`}{\color{`GREEN`}{`DATA_POINTS`}} \quad = \quad
\color{`BLUE`}{`MEAN`\text{ `plural( "year", MEAN )` old}}

Find the squared deviations from the mean for each sample. Since we don't know the
population mean, estimate the mean by using the sample mean we just calculated
`(\color{`

.
`BLUE`}{\overline{x}} = \color{`BLUE`}{`MEAN`\text{ `plural( "year", MEAN )`}})

Age

`x_i`

Distance from the mean
`(x_i - \color{``BLUE`}{\overline{x}})

`(x_i - \color{``BLUE`}{\overline{x}})^2

`POINT`

`roundTo( 2, POINT - MEAN )`

`SQR_DEV[ i ]`

`^2`

Normally we can find the variance `(\color{red}{s^2})`

by averaging the
squared deviations from the mean. But remember we don't know the real
population mean—we had to estimate it by using the sample mean.

The age of any particular `animal( 1 )` in our sample is likely to be closer to the average age
of the `plural( DATA_POINTS, animal( 1 ) )` we sampled than it is to the average age
of all `plural( POPULATION, animal( 1 ) )` in the zoo.
Because of that, the squared deviations from the mean we calculated will
probably underestimate the actual deviations from the population mean.

To compensate for this underestimation, rather than simply averaging the squared deviations from the mean,
we total them and divide by `n - 1`

.

```
\color{red}{s^2} \quad = \quad
\dfrac{\sum\limits_{i=1}^{\color{
```

`GREEN`}{n}} (x_i - \color{`BLUE`}{\overline{x}})^2}{\color{`GREEN`}{n - 1}}

```
\color{red}{s^2} \quad = \quad
\dfrac{
```

`plus.apply( KhanUtil, $.map( SQR_DEV, function( x ) { return "\\color{purple}{" + x + "}"; }) )`}
{\color{`GREEN`}{`DATA_POINTS` - 1}}

```
\color{red}{s^2} \quad = \quad
\dfrac{\color{purple}{
```

`roundTo( 2, sum( SQR_DEV ) )`}}{\color{`GREEN`}{`DATA_POINTS - 1`}} \quad = \quad
\color{red}{`VARIANCE`\text{ `plural( "year", VARIANCE )`}^2}

As you might guess from the notation, the sample standard deviation `(\color{`

is
found by taking the square root of the sample variance `PINK`}{s})`(\color{red}{s^2})`

.

`\color{`

`PINK`}{s} = \sqrt{\color{red}{s^2}}

```
\color{
```

`PINK`}{s} = \sqrt{\color{red}{`VARIANCE`\text{ `plural( "year", VARIANCE )`}^2}} =
\color{`PINK`}{`STDDEV`\text{ `plural( "year", STDDEV )`}}

**
We can estimate that the average animal( 1 ) at the zoo is plural( MEAN, "year" ) old with a standard deviation
of plural( STDDEV, "year" ).
**