Standard Deviation and Normal Distribution

• Jan 25th 2006, 09:17 AM
Arbitur
Standard Deviation and Normal Distribution
Hi I know how to calculate standard deviation but am having trouble with questions the in the image. Please help.. Either 8 or 10 worked out would be a huge help.
• Jan 25th 2006, 11:53 AM
ThePerfectHacker
Since 173 is the mean and standard deviation is 7.7 cm. Thus, One standard deviation to the left and right takes 68% (a rule) thus, 165.3-180.7 take 68% of all the students. Thus 32% below 165.3 or above 180.7 cm. Now the normal distribution is symettric thus, 16% below 165.3 and 16% aboce 180.7 cm. Look at the problem it say the taller 15% thus, approximately the taller 16% thus the range is approximately 180.7 cm.
• Jan 25th 2006, 11:57 AM
ThePerfectHacker
Look at the hint to look for the lower part of normal distribution. This problem unlike the first requires you to move 2 standard deviations. This will take on 95% (a rule) thus, between .9-4.1 years are included in these 95%. Thus, 5% below .9 or above 4.1 years. But because the normal distribution is symettric thus, 2.5% below .9 years or above 4.1 years. The problem asks for the 2.5% thus the range is .9 years.
• Jan 25th 2006, 01:37 PM
Arbitur
thanks... I know what your saying but could you just show how you calculated just one of those if possible? Thanks.
• Jan 25th 2006, 02:06 PM
ThePerfectHacker
Let $\displaystyle P(a\leq x\leq b)$ be the precentage (or probability) of all values between $\displaystyle a$ and $\displaystyle b$.

There is a theorem (theorem-1) in statistics that if something is normal distributed and $\displaystyle \bar x$ is the mean, and $\displaystyle \sigma$ is the standard deviation then,
$\displaystyle P(\bar x-\sigma\leq x\leq \bar x+\sigma)=.68$
$\displaystyle P(\bar x-2\sigma\leq x\leq \bar x+2\sigma)=.95$.
Also because (theorem-2) the normal distribution has symettrism we have that If $\displaystyle P(\bar x-a\leq x\leq \bar x+a)=p$ Then, $\displaystyle P(x\leq \bar x-a)=P(\bar x+a\leq x)=\frac{1-p}{2}$. The reason being, because $\displaystyle 1-p$ excludes the first one and then diving by 2 is because of symettrism.

Now to do problem 2, the mean $\displaystyle \bar x=2.5$ and the standard deviation is $\displaystyle \sigma=.8$. By the above explanations and theorem-1 $\displaystyle P(.9\leq x\leq 4.1)=.95$ (because 2 standard deviations). Next, use theorem-2 thus, $\displaystyle P(x\leq .9)=\frac{1-.95}{2}=.025$=2.5% which you where trying to find.
Thus the answer is $\displaystyle x\leq .9$.
Q.E.D.
• Jan 25th 2006, 02:09 PM
Arbitur
Arr.. thanks for trying but I still don't get how you got to the answer. I'll try a tutor somewhere for 1 session or something. (Taking this course online.) Thanks anyways.
• Jan 25th 2006, 02:23 PM
Arbitur
Heh.. feel stupid.. cant even figure out how you did it. Thanks neways :) guess im just not smart enough

Shouldnt it be 17% instead of 16%?
• Jan 25th 2006, 02:59 PM
ThePerfectHacker
Quote:

Originally Posted by Arbitur
Heh.. feel stupid.. cant even figure out how you did it. Thanks neways :) guess im just not smart enough

Shouldnt it be 17% instead of 16%?

How? 100-68=32 and 32/2=16
• Jan 27th 2006, 10:48 AM
CaptainBlack
Quote:

Originally Posted by Arbitur
Hi I know how to calculate standard deviation but am having trouble with questions the in the image. Please help.. Either 8 or 10 worked out would be a huge help.

8. If the school has a normal distribution of height $\displaystyle h$, with mean $\displaystyle 173 cm$ and
standard deviation $\displaystyle 7.7 cm$, then:

$\displaystyle z=\frac{h-173}{7.7}$,

has a standard normal distribution. The height of the beams is such that
$\displaystyle 15 \%$ will hit their heads, call this height $\displaystyle h_c$.

Call the corresponding $\displaystyle z$ score $\displaystyle z_c$, then:

$\displaystyle z_c=\frac{h_c-173}{7.7}$,

So $\displaystyle 15 \%$ of $\displaystyle z$ scores exceed $\displaystyle z_c$, or $\displaystyle 85 \%$ of $\displaystyle z$ scores are less than $\displaystyle z_c$.

Now look up the value of $\displaystyle z_c$ which gives $\displaystyle p(z<z_c)=0.85$ in a table
of the cumulative normal distribution, doing this gives $\displaystyle z_c \approx 1.036$.

The corresponding height (which will be the height of the beams) will be
the solution of:

$\displaystyle z_c = \frac{h_c - 173}{7.7}$,

which when we plug in the value of $\displaystyle z_c$ gives:

$\displaystyle h_c=180.98$.

Thus all students taller than $\displaystyle 180.98$ cm. will be affected.

RonL