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.
- Calculate the probability that the weight of a bag will be greater than $45$ kg.
- Calculate the probability that the weight of a bag will be less than $45$ kg.
- 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?
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
I think the answers should be 100% minus the answer from point a)
c)I have no idea :S
Can you help me?