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I am really confused by the following problem and I am not sure if my solution is correct.

Problem: In a shop there are 300 products which have come from 3 factories (A, B and C). The products from factory A are 70%, the products from factory B are 10%, from factory C are 20%. From the products produced in factory A 80% are first class. From the products produced in factory B only 40% are first class and from the products produced in factory C 75% are first class.

a) What is the probability that an arbitrary chosen product is first class?

b) What is the mean value and the standard deviation of the number of first class objects in the shop?

c) Find the probability that the number of first class products is between 210 and 225.

My solutions.

a) The total number of first class products is 0.8*0.7*300+0.4*0.1*300+0.2*0.75*300=225. So the probability that an arbitrary product is first class is 225/300=0.75.

b) Let $X_i$ be the number of first class products from the i-th factory, i=1,2, 3. Let $X =X_1+X_2+X_3$ be the number of first class products in the shop.

Then

$E(X)=E(X_1+X_2+X_3)=E(X_1)+E(X_2)+E(X_3)= 210*0.8+30*0.4+60*0.75=225 $ (because we have independent Bernoulli trials and the mean is np).

$V(X)= V(X_1)+V(X_2)+V(X_3) = 210*0.8*0.2 +30*0.4*0.6+60*0.75*0.25= 52.05$

and hence the standard deviation is $\sqrt(52.05)$.

Are my solutions to a) and b) correct?

How can I solve c)?

Thank you.

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  • $\begingroup$ You solutions for a) and b) look correct. For c) I think you can approximate it by the normal distribution. Let $X_1+X_2+X_2=S_3$ Applying the central limit distribution we get $P(210\leq S_3\leq 225)\approx \Phi\left(\frac{225-225}{\sqrt{52.05}} \right)-\Phi\left(\frac{209-225}{\sqrt{52.05}} \right)$ $\endgroup$ – callculus Mar 21 '19 at 21:18
  • $\begingroup$ Thanks. Why $\Phi (\frac{225-225}{\sqrt52.05})$? It becomes $\Phi(0)$ in this case. $\endgroup$ – Yuri Adamov Mar 21 '19 at 21:37
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    $\begingroup$ Because $P(S_3\leq 225)= \Phi (\frac{225-225}{\sqrt{52.05}})$. Indeed it becomes $\Phi(0)$. For a discrete random variable we have $P(x_1\leq X\leq x_2)=P(X\leq x_2)-P(X\leq x_1-1)$ $\endgroup$ – callculus Mar 21 '19 at 21:40
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The problem, as given, is not well-stated.

If you have "$300$ products" of which "the products from factory $A$ are $70\%$" , etc ... then we have a fixed number of products from $A$ ($0.7 \times 300=210$ ) , ... and the same goes for "first class products" and so on. But then question $b)$ makes little sense, because the "number of first class objects in the shop" is a fixed number, not a random variable.

So, I guess the statement means: we have picked 300 products, each of which is assumed to come from one of the factories $A,B,C$ with respective probabilities $(0.7,0.1,0.2)$ , and each of which can be first or second class with given probabilities (also, the products are independent). In this interpretation, the assertion "the products from factory $A$ are $70\%$" actually means "in average".

But, then, if this is so, your answer to $a)$ is wrong, because it assumes a fixed number of products of each factory and class.

The correct way is to write the given data as probabilities. Let $F \in \{A,B,C\}$ be the random variable that corresponds to the factory, and $Q\in \{1,2\}$ (quality) the class of a product. Then we know

$$P(F=A)=0.7 $$ $$P(F=B)=0.1 $$ $$P(F=C)=0.2 $$

$$P(Q=1 \mid F=A)=0.8 $$ $$P(Q=1 \mid F=B)=0.4 $$ $$P(Q=1 \mid F=C)=0.75 $$

Only now we try to answer the questions:

For $a)$:

$$P(Q=1) = \sum_F P(Q=1,F) = \sum_F P(Q=1 \mid F) P(F) =\\ =P(Q=1 \mid F=A)P(F=A) +P(Q=1 \mid F=B)P(F=B) +P(Q=1 \mid F=C)P(F=C) $$

Cna you go on from here?

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