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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 can’t understand c) I am not getting how the percentage between the mean and 7.985 kg after reduced is 48%? and how the answer has a z score of -2.05? please help
You have to work backwards using the tables provided in the exam (as I explain in my free lectures).
48% is 0.48. If you look in the tables along the 2.0 row then you will fine that for 2.05 we get 0.4798 and for 2.06 we get 0.4803. So the answer is nearest to 2.05.
Have you watched my free lectures? They are a complete free course for Paper MA and cover everything needed to be able to pass the exam well.
How did we get 48%? the z score for 7.985kg is -0.75 and so the percentage between the mean and 7.985kg is 0.2734 (27.34%) and it says percentage of items too light is below 7.985kg so the way I worked is, the percentage for too light weight item is = 0.5 – 0.2734 and from this I’m getting 0.2266 for too light weight items. Please tell me what I am doing wrong in this part Sir. And yes Sir, I have watched all of your free lectures
Please read my previous answer again carefully.
For the probability of being too light to be 2%, then the probability of being between the mean and the limit must be 50% – 2% = 48% (because of the fact that the curve is symmetrical).
Ohhh I get it now. I misunderstood what the question meant, I assumed it meant to reduce “by 2%” but it is actually “to 2%” 😀 Thankyou Sir for your help, I’m so relieved
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