Question 1092412: at medicare factory the amount which go into bottles of cough syrup are supposed to be normally distributed with mean 36 oz and standard deviation 0.1 oz.once avery 30 minutes a bottle is selected from the production line and its contents are noted precisely. if the amount of the bottle goes below 35.8oz or above 36.2oz,then the bottle will be declared out of control. if the process is in a control,mean is equal to 36 oz and standard deviation 0.1 oz
a).find the probability that abottle will be declared out of control
b)in asituation of a),find the probability that the number of bottles found out of control in an eight hour day(16 inspections )will be zero.
c)in situation of a,find the prbability that the number of bottles found out of control in an eight hour day (16inspectios) wll be exactly one
d) if the process shifts so that mean is equql to 37oz and standard deviation is 0.4 oz,find probability that abottle will be declared out of control
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
each x is +/- 0.2 when the mean is 36 and divided by the sd of 0.1, z is between -2 and 2, which has probability of in control is 0.9545.
P(out of control)=0.0455.
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b. That would be 0.9545^16=0.4747.
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c. This would be 16 ways to be 1 and 15 ways to be 0 so 16*0.0545*0.9545^15=0.3620
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d. To be out of control I am assuming the bottle should be between 35.8 and 36.2 still
that would be a z of (35.8-37)/0.4=-3 to (36.2-37)/0.4=-2. That would be .0214 probability in control and 0.9786 out of control. I'm not sure what the new limits are.
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