1. ## Probability question

The diameter of an orange is normally distributed with mean 7cm and standard deviation 1cm. Using a series of sieves, we wish to sort the oranges into three classes. Class 1 should contain large oranges, class 2 should contain medium oranges, and class 3 should contain small oranges in the proportion of 20%, 50%, and 30%.

Q1) How should we choose the gaps d1 and d2?
Q2) If the gaps are appropriately choses, what is the probability that out of 10 oranges, 6 or more oranges fall into class 2?

2. Originally Posted by woody198403
The diameter of an orange is normally distributed with mean 7cm and standard deviation 1cm. Using a series of sieves, we wish to sort the oranges into three classes. Class 1 should contain large oranges, class 2 should contain medium oranges, and class 3 should contain small oranges in the proportion of 20%, 50%, and 30%.

Q1) How should we choose the gaps d1 and d2?
Q2) If the gaps are appropriately choses, what is the probability that out of 10 oranges, 6 or more oranges fall into class 2?
a) Let $\displaystyle d_1, d_2$ be the desired diameters for class 1 and class 2 oranges, respectively, and $\displaystyle X$ the diameter of a randomly selected orange (random variable).
For $\displaystyle d_1$ we must have

$\displaystyle \begin{array}{lcll} \mathrm{P}(d_1\leq X) &=&0.2\\ 1-\mathrm{P}(X\leq d_1) &=& 0.2\\ \mathrm{P}(X\leq d_1) &=& 0.8\\ \mathrm{P}\left(\frac{X-7}{1}\leq \frac{d_1-7}{1}\right) &=& 0.8 &\big| \Phi^{-1}\\ d_1-7 &=& \Phi^{-1}(0.8)\\ d_1 &=& 7+\Phi^{-1}(0.8)\approx 7.84\mathrm{cm} \end{array}$

Similarly, for $\displaystyle d_2$ we must have that
$\displaystyle \begin{array}{lcll} \mathrm{P}(d_2\leq X) &=&0.2+0.5\\ 1-\mathrm{P}(X\leq d_2) &=& 0.7\\ \mathrm{P}(X\leq d_2) &=& 0.3\\ \mathrm{P}\left(\frac{X-7}{1}\leq \frac{d_2-7}{1}\right) &=& 0.3 &\big| \Phi^{-1}\\ d_2-7 &=& \Phi^{-1}(0.3)\\ d_2 &=& 7+\Phi^{-1}(0.3)\approx 6.45\mathrm{cm} \end{array}$

b) Let $\displaystyle Y$ be the number of Class 2 oranges in a selection of $\displaystyle n=10$ oranges. Since the probability of a single, randomly selected orange to be in Class 2 is $\displaystyle p=0.5$ we get

$\displaystyle \mathrm{P}(6\leq Y) = \sum_{y=6}^{10}\binom{10}{y} 0.5^y\cdot (1-0.5)^{10-y}=\sum_{y=6}^{10}\binom{10}{y} 0.5^{10}\approx 0.38=38\%$

3. Thanks Failure,
but I am a little confused with what you have written for part (a) for d2.

Should it be
$\displaystyle 1 - P(X \leq d_2) = 0.7$
$\displaystyle P(X \leq d_2) = 0.3$

or am I not looking at it the right way?

Also, can you please explain why you have used $\displaystyle \Phi^{-1}$?
Isn't it the symbol for the pdf of a normal distribution?

4. Originally Posted by woody198403
Thanks Failure,
but I am a little confused with what you have written for part (a) for d2.

Should it be
$\displaystyle 1 - P(X \leq d_2) = 0.7$
$\displaystyle P(X \leq d_2) = 0.3$

or am I not looking at it the right way?
No, you are absolutely right: that was a terrible typo - I am going to modify that answer right away...

Also, can you please explain why you have used $\displaystyle \Phi^{-1}$?
I had to get rid of $\displaystyle \mathrm{P}(\ldots)$ on the left side of the equation in order to be able to isolate $\displaystyle d_1-7$ and $\displaystyle d_2-7$ respectively.
Isn't it the symbol for the pdf of a normal distribution?
Not quite. I use the definition $\displaystyle \Phi(x) := \mathrm{P}(X\leq x)$ for a normally distributed random variable X with mean 0 and variance 1. So $\displaystyle \Phi$ is the pdf (more precisely, the cumulative distribution function) of a normally distributed random variable with mean 0 and variance 1 ("standard normal cdf"). And $\displaystyle \Phi^{-1}$ is simply the inverse of that (monotonously increasing) function $\displaystyle \Phi:\mathbb{R}\to (0,1)$ (also called the quantile function of a normally distributed random variable). You get it by simply reading a table for $\displaystyle \Phi$ the other way around...