1. ## [SOLVED] Normal Distribution

Question:
The length of sweetpea flower stems are normally distrusted with mean $18.2 cm$ and standard deviation $2.3 cm$.
(a) Find the probability that the length of a flower stem is between $16 cm$ and $20 cm$.
(b) $12\%$ of the flower stems are longer than h cm. $20\%$ of the flower stems are shorter than k cm. Find $h$ and $k$.
(c) Stem lengths less than $14 cm$ are unacceptable at a florist's shop. In a batch of $500$ sweetpeas estimate how many would be unacceptable.

Attempt:

$\mu = 18.2$ , $\sigma = 2.3$

$(a)$ $P( 16 < X < 20 )$

$P \left( \frac{(16 -18.20}{2.3} < \frac{X - \mu}{\sigma} < \frac{(20-18.2)}{2.3} \right)$

$P( -0.957 < Z < 0.783)$
$= \phi(0.783) - ( 1 - \phi(0.8289 + 0.0018))$
$= 0.6139$

$(b)$ Don't know how to start, need help!

2. Originally Posted by looi76
Question:
The length of sweetpea flower stems are normally distrusted with mean $18.2 cm$ and standard deviation $2.3 cm$.
(a) Find the probability that the length of a flower stem is between $16 cm$ and $20 cm$.
(b) $12\%$ of the flower stems are longer than h cm. $20\%$ of the flower stems are shorter than k cm. Find $h$ and $k$.
(c) Stem lengths less than $14 cm$ are unacceptable at a florist's shop. In a batch of $500$ sweetpeas estimate how many would be unacceptable.

Attempt:

$\mu = 18.2$ , $\sigma = 2.3$

$(a)$ $P( 16 < X < 20 )$

$P \left( \frac{(16 -18.20}{2.3} < \frac{X - \mu}{\sigma} < \frac{(20-18.2)}{2.3} \right)$

$P( -0.957 < Z < 0.783)$
$= \phi(0.783) - ( 1 - \phi(0.8289 + 0.0018))$
$= 0.6139$

$(b)$ Don't know how to start, need help!
Well if your answer was off by .0001 I think you can safely assume you are correct

3. Originally Posted by Mathstud28
Well if your answer was off by .0001 I think you can safely assume you are correct
I need help in the second step!... I don't know how to start

4. Hi

You want to find $h=\phi^{-1}(88\%)$ which is exactly what you had to do in another exercise involving the life time of an object. (substitute $u=\frac{x-\mu}{\sigma}$ in the integral and solve)

5. Originally Posted by flyingsquirrel
Hi

You want to find $h=\phi^{-1}(88\%)$ which is exactly what you had to do in another exercise involving the life time of an object. (substitute $u=\frac{x-\mu}{\sigma}$ in the integral and solve)
So is $\phi^{-1}(88\%) = .8800$
$\phi = 8106$

Is this step right?

6. I don't understand why you picked up .8106 instead of .8810 ?

Having done the substitution, we are looking for $h$ such that $\int_{-\infty}^{\frac{h-\mu}{\sigma}} \frac{1}{\sqrt{2\pi}}\exp{-u^2}\,\mathrm{d}u=0.88$ and the table gives $\frac{h-\mu}{\sigma}=1.18$ for 0.8810, doesn't it ?

7. Originally Posted by looi76
So is $\phi^{-1}(88\%) = .8800$
$\phi = 8106$

Is this step right?
$z = \frac{x - \mu}{\sigma}$.

Pr(X > h) = 0.12 => Pr(X < h) = 0.88.
Use your tables to find a value $z_{0.88}$ of z corresponding to $\Pr(z < z_{0.88}) = 0.88$: $z_{0.88} = 1.175$. Therefore h = .......

Pr(X < k) = 0.2 => Pr(X > -k) = 0.2 => Pr(X < -k) = 0.8.
Use your tables to find a value $z_{0.8}$ of z corresponding to $\Pr(z < z_{0.8}) = 0.8$: $z_{0.8} = 0.842$. Therefore k = .......

8. Originally Posted by flyingsquirrel
I don't understand why you picked up .8106 instead of .8810 ?

Having done the substitution, we are looking for $h$ such that $\int_{-\infty}^{\frac{h-\mu}{\sigma}} \frac{1}{\sqrt{2\pi}}\exp{-u^2}\,\mathrm{d}u=0.88$ and the table gives $\frac{h-\mu}{\sigma}=1.18$ for 0.8810, doesn't it ?
Originally Posted by mr fantastic
$z = \frac{x - \mu}{\sigma}$.

Pr(X > h) = 0.12 => Pr(X < h) = 0.88.
Use your tables to find a value $z_{0.88}$ of z corresponding to $\Pr(z < z_{0.88}) = 0.88$: $z_{0.88} = 1.175$. Therefore h = .......

Pr(X < k) = 0.2 => Pr(X > -k) = 0.2 => Pr(X < -k) = 0.8.
Use your tables to find a value $z_{0.8}$ of z corresponding to $\Pr(z < z_{0.8}) = 0.8$: $z_{0.8} = 0.842$. Therefore k = .......
Thanks flyingsquirrel and mr fantastic

$= P( h \geq 12\% )$
$= P( h \geq -88\% )$████ $12\% - 100\% = -88\%$
$= \phi^{-1}(88\%)$
$= \phi^{-1}(0.8800)$
$= 1.175$

$= P( k \leq 20\% )$
$= P( k \leq -80\% )$████ $20\% - 100\% = -80\%$
$= \phi^{-1}(-80\%)$
$= \phi^{-1}(1- 0.8000)$
$= 1 - 0.842$
$= 0.158$

In textbook answer for h is 20.9 and k is 16.3 ?

9. Originally Posted by looi76
$P( h \geq 12\% )$
What does this mean ? The probability that the length $h$ is greater than 12% ? Writing $P(X\geq h)=12\%$ would be OK : "the probability that the length of a stem is greater than $h$ equals 12%"

You'd better written what Mr Fantastic suggested :
$P(X>h)=12\%$ hence $P(x
$\frac{h-\mu}{\sigma}=\phi^{-1}(88\%)=1.18$
Whence $h=1.18\times2.3+18.2=20.9\,\mathrm{cm}$

I let you redo the second one accordingly.
In textbook answer for h is 20.9 and k is 16.3 ?
Yes...

10. Originally Posted by mr fantastic
$z = \frac{x - \mu}{\sigma}$.

Pr(X > h) = 0.12 => Pr(X < h) = 0.88.
Use your tables to find a value $z_{0.88}$ of z corresponding to $\Pr(z < z_{0.88}) = 0.88$: $z_{0.88} = 1.175$. Therefore h = .......

Mr F continues: ${\color{red}z = \frac{x - \mu}{\sigma} \Rightarrow 1.175 = \frac{h - 18.2}{2.3} \Rightarrow h = 20.90}$ (correct to two decimal places).

You should check that Pr(X > 20.9) = 0.12 ......

Pr(X < k) = 0.2 => Pr(X > -k) = 0.2 => Pr(X < -k) = 0.8.
Use your tables to find a value $z_{0.8}$ of z corresponding to $\Pr(z < z_{0.8}) = 0.8$: $z_{0.8} = 0.842$. Therefore k = .......
The other is done in the same way.