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The manufacturer A produces different types of cement bags, their weight (in kg) is a normal random variable with mean of $45$ kg and deviation of $4$ kg.

The competitor B produces the same type of cement, their weight has mean of $41$kg and deviation of $12$ kg, but in this case it is a uniform distribution (not normal). A site manager buys $40\%$ cement of A and the rest from B.

  1. Calculate the probability that the weight of a bag will be greater than $45$ kg.
  2. Calculate the probability that the weight of a bag will be less than $45$ kg.
  3. The site manager must carry $10$ bags of cement from A and 90 bags of cement from B. What is the probability that the total load will not exceep the specified maximum value of $4$ tons?

a)(NOT SURE)

The standard deviation of a uniform distribution is (b-a)/sqrt(12), where b and a are the maximum and minimum values that can be assumed by the variable, right?

If the standard deviation of the cement from B is 12 kg, the difference between the maximum and minimum values of the cement bags from B is 12*sqrt(12) = 41.569 kg. Thus, the weights of bags from B are uniformly distributed over the range 20.72 kg to 61.28 kg.


The probability that a bag from A will weigh more than 45 kg is 50%, and the probability that a bag from B will weigh more than 45 kg is: (61.28 - 45)/41.569

So, answer should be: 0.40 * 0.50 + 0.60 * (61.28 - 45)/41.569 = 0.43498


b)(NOT SURE)

I think the answers should be 100% minus the answer from point a)


c)I have no idea :S


Can you help me?

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  • $\begingroup$ (a) and (b) seem correct to me. $\endgroup$ – gt6989b Jul 13 '17 at 16:27
  • $\begingroup$ Thank you for replying! What about point c)? :S $\endgroup$ – JustToKnow Jul 13 '17 at 17:03
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1+2 indeed right. It is the law of total probability, i.e., \begin{align} P(X<45) &= P(X<45|X=X_A)P(X=X_A)+P(X<45|X=X_B)P(X=X_B)\\ &= 0.4\times 0.5 + 0.6 \times \frac{45-a}{41}. \end{align}

  1. For a exact answer, assuming that all the r.vs are independent, you will need to find the distribution of $W = \sum_{i=1}^{90} X_i\sim N(90\times 45, 4^2\times 45)$ $+$ $V = \sum_{j=1}^{10} U_i$ that is "shifted" Irwin-Hall distribution. I can guess that it will not have an analytical closed form. Thus, you can you use the advice of @gt6989b and bound the probability with Chebyshev's inequality.

$$ Var(W+V)= 4^2\times 45 + 12^2\times 10, $$ hence, $$ P(|W+V - (41 \times 10 + 45\times 90)|\ge 4000)\le \frac{4^2\times 45 + 12^2\times 10}{4000^2} = 0.54. $$

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(a) and (b) are correct. I do not see a good exact solution for (c), but you can bound it.

You have to compute $\mathbb{P}[X < 4000]$ where $X = A_1 + \ldots + A_{10} + B_{1} + \ldots B_{90}$, which is a nasty 100-dimensional integral.

Since you know $\mu_A, \mu_B, \sigma_A^2, \sigma_B^2$, you can compute $\mu_X$ and $\sigma_X^2$ to plug into Chebyshev's inequality, $$ \mathbb{P}\left[|X- \mu_X| > k\sigma_X^2\right] \le \frac{1}{k^2} $$ and Markov's inequality $$ \mathbb{P}[X > a] \le \frac{\mu_X}{a} $$

For example, since all $a_k, B_k$ can be assumed independent, $$ \mu_X = \mathbb{E}\left[\sum_{k=1}^{10}A_k + \sum_{k=1}^{90}B_k + \right] = 10 \mu_A + 90 \mu_B $$ and you can find $\sigma_X^2$ this way as well...

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  • $\begingroup$ I have tried but i do not understand how to do the point c) :S $\endgroup$ – JustToKnow Jul 13 '17 at 20:22

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