Please Help Me!

**crazy_gal108** · March 10th 2006, 07:06 AM

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.

**ThePerfectHacker** · March 10th 2006, 08:40 AM

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,
$\sqrt{\frac{x_1^2+x_2^2+...+x_n^2}{n}$
The formula for variance is,
$\frac{x_1^2+x_2^2+...+x_n^2}{n}$
Thus, we have,
$\sqrt{\frac{x_1^2+x_2^2+...+x_n^2}{n}}> \frac{x_1^2+x_2^2+...+x_n^2}{n}\geq 0$
Square both sides,
$\frac{x_1^2+...+x_n^2}{n}>\frac{(x_1^2+...+x_n^2)^ 2}{n^2}$
Thus, since $n>0$ , let $y=x_1^2+...+x_n^2$
To get,
$ny>y^2$
Since, $y>0$ divide by it to get,
$n>y$
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.

**ThePerfectHacker** · March 10th 2006, 08:47 AM

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 $z\approx 1.96$ . Thus, the upper range is $15+(1.96)(.1)=15.196$ and the lower range is $15-(1.96)(.1)=14.804$ .
Thus, the range is
$14.804\leq x\leq 15.196$

**CaptainBlack** · March 10th 2006, 12:18 PM

Originally Posted by ThePerfectHacker

The formula for standard deviation is,
$\sqrt{\frac{x_1^2+x_2^2+...+x_n^2}{n}$
The formula for variance is,
$\frac{x_1^2+x_2^2+...+x_n^2}{n}$

The formula for the variance of a sample is:

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

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

The standard deviation is:

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

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

RonL

**ThePerfectHacker** · March 11th 2006, 03:12 PM

Originally Posted by CaptainBlack

The formula for the variance of a sample is:

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

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

The standard deviation is:

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

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

RonL

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