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Question
The weights of a certain mass-produced item are known, over a long period of time, to be normally distributed with a mean of 8 kg and a standard deviation of 0.02 kg.
Required:
(a) If items whose weight lies outside the range 7.985 – 8.035 kg are deemed to be faulty, what percentage of products will be faulty?
(b) If it is required to reduce the percentage of items that are too heavy (with weight over 8.035 kg) to 2%, to what value must the mean weight be decreased, leaving all other factors unchanged?
(c) If it is required to reduce the percentage of items that are too light (with weight below 7.985 kg) to 2%, to what value must the standard deviation be decreased, leaving other factors unchanged?Sir i dont get part b in answer it says
The tail-end probability of 2% corresponds to the table entry 48%
= 0.48, and so to the z-value 2.05.
Use the formula to work back to calculate the mean: 2.05 = (8.035 – ?)/0.02I dont get the tail end probability how they get 48% part can you explain
Because the normal distribution is symmetrical, the % with a weight of more than the mean must be 50%.
If 2% weigh over 8.035kg, then the % weighing between the mean and 8.035kg must be 50 – 2 = 48% or 0.48.It seems as though you are happy with the rest of the workings, but if not then ask again.
Okay sir i think now i get it what question was exactly saying i did with diagram again and with your explanation i am getting it now somewhat please keep this comment open if i get doubt i will ask in this related to this question
Sir you can close it i cleared my exam, thanks alot for videos your explanation are easy and good i really like the open tuition lookinf forward for next exam
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