1) Fred was told by his teacher that the mean final mark of all the students in his class was exactly 68.7. He surveyed these students and recorded the results in a frequency table with intervals 1-10, 11-20, etc. Will his calculated mean equal the one calculated by his teacher? Explain why or why not.

2) Explain what information is given by the magnitude and sign of the z-score of a piece of data.

3) When is the standard deviation larger than the variance? Explain with an example.

4) The lengths, in centimetres, of wooden pencils are normally distributed with a mean of 15 and a standard deviation of 0.1. Find two intervals of pencil lengths that will each include 81.5% of pencils.

2. Originally Posted by crazy_gal108

3) When is the standard deviation larger than the variance? Explain with an example.
The formula for standard deviation is,
$\displaystyle \sqrt{\frac{x_1^2+x_2^2+...+x_n^2}{n}$
The formula for variance is,
$\displaystyle \frac{x_1^2+x_2^2+...+x_n^2}{n}$
Thus, we have,
$\displaystyle \sqrt{\frac{x_1^2+x_2^2+...+x_n^2}{n}}> \frac{x_1^2+x_2^2+...+x_n^2}{n}\geq 0$
Square both sides,
$\displaystyle \frac{x_1^2+...+x_n^2}{n}>\frac{(x_1^2+...+x_n^2)^ 2}{n^2}$
Thus, since $\displaystyle n>0$, let $\displaystyle y=x_1^2+...+x_n^2$
To get,
$\displaystyle ny>y^2$
Since, $\displaystyle y>0$ divide by it to get,
$\displaystyle n>y$
Thus, the necessary conditions are when the sum of squares are less than the number of terms.

Checking this inequality we also find that this are the sufficient conditions.
Q.E.D.

3. Originally Posted by crazy_gal108

4) The lengths, in centimetres, of wooden pencils are normally distributed with a mean of 15 and a standard deviation of 0.1. Find two intervals of pencil lengths that will each include 81.5% of pencils.
I assume you want you solution centered at the mean.

Since, 81.5% of the area under the normal curve, we have that 40.75% of area of each side of the curve. Looking at the distribution tables,
http://www.statsoft.com/textbook/sttable.html
We find that for $\displaystyle z\approx 1.96$. Thus, the upper range is $\displaystyle 15+(1.96)(.1)=15.196$ and the lower range is $\displaystyle 15-(1.96)(.1)=14.804$.
Thus, the range is
$\displaystyle 14.804\leq x\leq 15.196$

4. Originally Posted by ThePerfectHacker
The formula for standard deviation is,
$\displaystyle \sqrt{\frac{x_1^2+x_2^2+...+x_n^2}{n}$
The formula for variance is,
$\displaystyle \frac{x_1^2+x_2^2+...+x_n^2}{n}$
The formula for the variance of a sample is:

$\displaystyle s^2=\frac{1}{n}\sum_1^n(x_i-\bar x)^2$

where $\displaystyle \{x_1,\ .. \ ,x_n\}$ is the sample of size $\displaystyle n$,
and $\displaystyle \bar x$ is the sample mean.

The standard deviation is:

$\displaystyle s=\sqrt{s^2}$.

Thus the standard deviation is less than the variance when the variance is
less than one.

RonL

5. Originally Posted by CaptainBlack
The formula for the variance of a sample is:

$\displaystyle s^2=\frac{1}{n}\sum_1^n(x_i-\bar x)^2$

where $\displaystyle \{x_1,\ .. \ ,x_n\}$ is the sample of size $\displaystyle n$,
and $\displaystyle \bar x$ is the sample mean.

The standard deviation is:

$\displaystyle s=\sqrt{s^2}$.

Thus the standard deviation is less than the variance when the variance is
less than one.

RonL
Oops,