# Thread: Standard Deviation and Normal Distribution

1. ## 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.

2. 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.

3. 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.

4. thanks... I know what your saying but could you just show how you calculated just one of those if possible? Thanks.

5. 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.

6. 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.

7. 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%?

8. 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

9. 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