In question three, if the factory decides to only limit the percentage of cement

Question

In question three, if the factory decides to only limit the percentage of cement with excessive gypsum content to 2%, what will be the percentage of produced cement having gypsum content less than the lower specification limit?

I NEED THE ANSWER TO THE QUESTION I LISTED ABOVE(#4 ON THE PICTURE). I AM ONLY POSTING THE OTHER QUESTIONS BECAUSE THEY ARE PART OF #4

2- 020 pts A cement manufacturing factory is trying to control the gypsum content in the produced cement in the range of 4.5 to 5.5% these two numbers are the lower and upper specification limits). Note that the presence of gypsum in cement is necessary for controlling the setting time of cement. A long-term investigation on the gypsum content of the produced cement samples suggests that the gypsum content (G) in the final product is a random variable having a normal distribution with an average and standard deviation of 5.2% and 0.3%. Find out what proportion percentage) of the produced cement will have gypsum content outside the specification limits 3- (15 pts) In the previous question, if the factory decides to limit the percentage of cement with excessively high or low gypsum content to 3%, assuming the average stays constant (5.3%), what is the maximum allowable standard deviation of G? excessive gypsum content to 2%, what will be the percentage of produced cement having gypsum content less than the lower specification limit?

Explanation / Answer

4.) This is a good question containing a rather unusual application of both conditional probability and normal distribution but can be solved in the following manner:

Required probability= P(G<0.045P(G>0.055)=0.02). We will find out the standard deviation for which P(G>0.055)=0.02 as in part 3 then use this standard deviation to calculate P(G<0.045) given the mean is constant at0.052. As you did in part three do it and you will get an answer that s.d= 0014607

Using G~N(0.052,0.0014607) we fing probability that P(G<0.045). As we did in part 1 you get an answer as P(N(0,1))<-4.792 which is almost=0

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