Distrubition help needed!

**Hikaru** · May 13th 2008, 04:10 AM

I'm sorry for asking a maths question again. XD

Q8. The smallest 84% of stainless steel nails are less than 49.7mm long, and the largest 97.5% are longer than 41.6mm long.

a. Caculate the likely mean length of the nails and an appropriate value for the standard deviation.

What do I do? >.<

Thanks in advance! =]

**janvdl** · May 13th 2008, 05:00 AM

Originally Posted by Hikaru

I'm sorry for asking a maths question again. XD

Q8. The smallest 84% of stainless steel nails are less than 49.7mm long, and the largest 97.5% are longer than 41.6mm long.

a. Caculate the likely mean length of the nails and an appropriate value for the standard deviation.

What do I do? >.<

Thanks in advance! =]

This is a normal distribution.

We can say that the chances that a nail is shorter than 41.6mm is 2.5%
And the chances that a nail is longer than 49.7mm is 16%

$P(X < 41.6) = 0.025$

$P \left( \frac{X - \mu}{\sigma} < \frac{41.6 - \mu}{\sigma} \right) = 0.025$

$P \left( Z < \frac{41.6 - \mu}{\sigma} \right) = 0.025$

$\Phi \left( \frac{41.6 - \mu}{\sigma} \right) = 0.025$

$\frac{41.6 - \mu}{\sigma} = -1.96$

=====

$P(X > 49.7) = 0.16$

$1 - P(X \leq 49.7) = 0.16$

$P \left( \frac{X - \mu}{\sigma} \leq \frac{49.7 - \mu}{\sigma} \right) =$

$P \left( Z \leq \frac{49.7 - \mu}{\sigma} \right) = 0.84$

$\Phi \left( \frac{49.7 - \mu}{\sigma} \right) = 0.84$

$\frac{49.7 - \mu}{\sigma} = 0.996$

====

From here on it's simply a matter of setting those two equations equal to each other and solving.

**mr fantastic** · May 13th 2008, 05:10 AM

Originally Posted by Hikaru

I'm sorry for asking a maths question again. XD

Q8. The smallest 84% of stainless steel nails are less than 49.7mm long, and the largest 97.5% are longer than 41.6mm long.

a. Caculate the likely mean length of the nails and an appropriate value for the standard deviation.

What do I do? >.<

Thanks in advance! =]

Let X be the random variable length of nail and assume X ~ Normal( $\mu, ~ \sigma)$ .

$\Pr(X < 49.7) = 0.84 \Rightarrow \Pr\left( Z < \frac{49.7-\mu}{\sigma}\right) = 0.84$

But $\Pr(Z < a) = 0.84 \Rightarrow a \approx 0.99446$ .

Therefore:

$\frac{49.7-\mu}{\sigma} = 0.99446$ .... (1)

$\Pr(X > 41.6) = 0.975 \Rightarrow \Pr\left( Z > \frac{41.6-\mu}{\sigma}\right) = 0.975$

But $\Pr(Z > a) = 0.975 \Rightarrow a \approx -1.95996$ .

Therefore:

$\frac{41.6 -\mu}{\sigma} = -1.95996$ .... (2)

Now solve equations (1) and (2) simultaneously.

For checking purposes: I get $\mu = 46.97$ , correct to two decimal places.

**janvdl** · May 13th 2008, 05:11 AM

Originally Posted by mr fantastic

Let X be the random variable length of nail and assume X ~ Normal( $\mu, ~ \sigma)$ .

$\Pr(X < 49.7) = 0.84 \Rightarrow \Pr\left( Z < \frac{49.7-\mu}{\sigma}\right) = 0.84$

But $\Pr(Z < a) = 0.84 \Rightarrow a \approx 0.99446$ .

Therefore:

$\frac{49.7-\mu}{\sigma} = 0.99446$ .... (1)

$\Pr(X > 41.6) = 0.975 \Rightarrow \Pr\left( Z > \frac{41.6-\mu}{\sigma}\right) = 0.975$

But $\Pr(Z > a) = 0.975 \Rightarrow a \approx -1.95996$ .

Therefore:

$\frac{41.6 -\mu}{\sigma} = -1.95996$ .... (2)

Now solve equations (1) and (2) simultaneously.

Mr F, your distribution table seems to be more accurate than mine...

**mr fantastic** · May 13th 2008, 05:16 AM

Originally Posted by janvdl

This is a normal distribution.

We can say that the chances that a nail is shorter than 41.6mm is 2.5%
And the chances that a nail is longer than 49.7mm is 16%

$P(X < 41.6) = 0.025$

$P \left( \frac{X - \mu}{\sigma} < \frac{41.6 - \mu}{\sigma} \right) = 0.025$

$P \left( Z < \frac{41.6 - \mu}{\sigma} \right) = 0.025$

$\Phi \left( \frac{41.6 - \mu}{\sigma} \right) = 0.025$

$\frac{41.6 - \mu}{\sigma} = -1.96$

=====

$P(X > 49.7) = 0.16$

$1 - P(X \leq 49.7) = 0.16$

$P \left( \frac{X - \mu}{\sigma} \leq \frac{49.7 - \mu}{\sigma} \right) =$

$P \left( Z \leq \frac{49.7 - \mu}{\sigma} \right) = 0.84$

$\Phi \left( \frac{49.7 - \mu}{\sigma} \right) = 0.84$

$\frac{49.7 - \mu}{\sigma} = 0.996$

====

From here on it's simply a matter of setting those two equations equal to each other and solving.

That's a first. Getting beaten to the punch on a statistics question

**mr fantastic** · May 13th 2008, 05:17 AM

Originally Posted by janvdl

Mr F, your distribution table seems to be more accurate than mine...

And my TI-89 is even more accurate .....

**janvdl** · May 13th 2008, 05:18 AM

Originally Posted by mr fantastic

And my TI-89 is even more accurate .....

Ah okay I get it

My calculator isn't capable of doing that...

**mr fantastic** · May 13th 2008, 05:44 AM

Originally Posted by janvdl

Ah okay I get it

My calculator isn't capable of doing that...

There are many on-line calculators that can do it for you. For example:

z score Calculator

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