# Thread: Normal distribution question

1. ## Normal distribution question

My question is as follows: Speed driven by motorists are normally distributed with mean 63 and standard deviation unknown.

1.) If 2.5% of all motorists are traveling at speed 70 or more, what is the value for standard deviation?

2.) Mean and standard deviation are unknown. It is observed that 10% of all motorists travel at less than speed 36 and 10% of all motorists travel at more than speed 70. What must mean and standard deviation be?

2. ## Re: Normal distribution question

Originally Posted by romsek
My question is as follows: Speed driven by motorists are normally distributed with mean 63 and standard deviation unknown.

1.) If 2.5% of all motorists are traveling at speed 70 or more, what is the value for standard deviation?

2.) Mean and standard deviation are unknown. It is observed that 10% of all motorists travel at less than speed 36 and 10% of all motorists travel at more than speed 70. What must mean and standard deviation be?
$\Phi()$ below refers to the CDF of the standard normal distribution.

1) $P[s>70] = 0.025$

$1-P[s<70] = 0.025$

$P[s<70] = 0.975$

so we know that

$\Phi\left(\dfrac{70-63}{\sigma} \right) = 0.975$

Using a table, or software, or what have you, you then get

$\dfrac{7}{\sigma} \approx 1.96$

$\sigma \approx \dfrac{7}{1.96} = 3.57$

2) $P[s<36]=0.1,~P[s>70]=0.1$

well you have 2 equations now in $\mu$ and $\sigma$

$\Phi\left(\dfrac{36-\mu}{\sigma}\right) = 0.1$

$1-\Phi\left(\dfrac{70-\mu}{\sigma}\right) = 0.1$

apply the same table lookup done in (1) to get rid of $\Phi()$ and solve the resulting algebraic system.

I leave that to you.

3. ## Re: Normal distribution question

Essentially, this is a matter of being able to read a table of the "standard normal distribution". A good one is here: https://www.stat.tamu.edu/~lzhou/sta...ormaltable.pdf.

"2.5% of them are larger than this number" corresponds to probability 0.025 larger than the number so 1- 0.025= 0.975 corresponds to z= 1.96. That is, z is 1.96 standard deviations above the mean. Here, we are told that the mean is 63 so, taking the standard deviation to be "$\displaystyle \sigma$", $\displaystyle 70= 53+ 1.96\sigma$. Solve that equation for $\displaystyle \sigma$.

4. ## Re: Normal distribution question

Does anyone think that the OP was meant to use the 68-95-99.7 Approximation Rule?

5. ## Re: Normal distribution question

Originally Posted by Prove It
Does anyone think that the OP was meant to use the 68-95-99.7 Approximation Rule?
If the question asked for an approximation or rough figure, yes. Otherwise, it would drive me NUTS to make that assumption when 95 is closer to 95.5.

6. ## Re: Normal distribution question

Can someone please help me better understand #2 of the original question / provide the solution?

7. ## Re: Normal distribution question

Again, look at a table. The lowest 10% in a standard normal distribution is z= -2.3. That is, 2.3 and more standard deviations below the mean. $\displaystyle \frac{36- \mu}{\sigma}= -2.3$. The highest 10% is at z= 1.29 above the mean. $\displaystyle \frac{70- \mu}{\sigma}= 1.29$. Solve those two equations for $\displaystyle \mu$ and $\displaystyle \sigma$.

8. ## Re: Normal distribution question

For example, would I do 36/-2.3 to get the std and then find the mean accordingly from that? How do I combine the two (highest & lowest)?