In a factory there are $2$ machines that create tubes (They are independent from each other) Length of the tubes of machine A is distributed normally with an expectancy of $101$ cm. and a Variance of $102$. ($\mu = 101, V[A]=102$)
The tubes of the second machine is distributed uniformly $U(85, 115)$
A proper length of a tube is $[90,110]$, it is known that $90\%$ of all the tubes are fine.
- What is the percentage of the tubes created in Machine A ?
I've tried to solve it, but I get a weird answer:
$X \sim N(101, \sqrt{102})$ - Machine A (and $\sqrt{102}$ because $\sigma = \sqrt{\mathbb{V}[X]}$ )
$Y \sim \text{Uniform}(85,115)$ - Machine B
So we can build this equation:
$a \cdot \mathbb{P}(90 \leq X \leq 110) + (1-a) \cdot \mathbb{P}(90 \leq Y \leq 110) = 0.9$
Where $a$ is the percentage of all the tubes created at machine A (and thus $1-a$ is the percentage of machine B , however I am not sure if this is the right way of doing it0
I calculate $\mathbb{P}(90 \leq X \leq 110) = P(Z \leq 0.89) - P(Z \leq -1.09) = 0.6754$
And:
$\mathbb{P}(90 \leq Y \leq 110) = \frac{1}{115-85} \cdot (110-90) = \frac{2}{3}$
And I get:
$a \cdot 0.6754 + (1-a) \cdot \frac{2}{3} = 0.9$
But $a = 26.7$ which is greater than $1$. I am very confused what I did wrong here.
What I think the mistake is: I am not sure if the information that the machines are independent from each other mean that: $P_{A} + P_{B} = 1$
meaning, I am not sure that the percentage machine A creates + percentage of machine B creates = $1$ , but it seemed logical, no?
What is the problem?
Thank you!
Then to get a you have to solve
$$\frac{0.6755\cdot a}{ 0.6755\cdot a +(1-a)\cdot \frac{2}{3}}=0.9$$
leading to $a\approx 0.899$
I want to underline that "Variant" in Statistics has no a nice meaning.
We have "variance" and "standard deviation".
If you 102 is expressed in cm cannot be a variance but a standard deviation....so verify what your data mean....anyway the procedure does not change.