Each day a quality engineer selects a random sample of 60 power supplies

Poke

Homework Statement


Each day a quality engineer selects a random sample of 60 power supplies from the day's production, measures their output voltages, and computes a 90% confidence interval for the mean output voltage of all the power supplies manufactured that day. What is the probability that more than 15 of the confidence intervals constructured in the next 210 days will fail to cover the true mean? Hint: Use the normal approximation.

Homework Equations


z=\Frac{x-\miu}{\sigma}

The Attempt at a Solution


First, X~Bin(n,p), as np >10, it follows Normal distribution, X~N(np, np(1-p))

np is \miu,
\sqrt{np(1-p)} is \signma

then Z = \frac{x-\miu}{\sigma}, it is the area of left tail
 
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Poke said:

Homework Statement


Each day a quality engineer selects a random sample of 60 power supplies from the day's production, measures their output voltages, and computes a 90% confidence interval for the mean output voltage of all the power supplies manufactured that day. What is the probability that more than 15 of the confidence intervals constructured in the next 210 days will fail to cover the true mean? Hint: Use the normal approximation.

Homework Equations


z=\Frac{x-\miu}{\sigma}

The Attempt at a Solution


First, X~Bin(n,p), as np >10, it follows Normal distribution, X~N(np, np(1-p))

np is \miu,
\sqrt{np(1-p)} is \signma

then Z = \frac{x-\miu}{\sigma}, it is the area of left tail

If ##X\sim\text{Bin}(n,p)##, the statement that ##X## "follows a normal distribution" when ##np > 10## is patently false: it may be approximately normal, at best. Furthermore, an ##np## of about 10 is too small for the normal to be really accurate, but using the "1/2" correction can improve the approximation quite a bit. Finally: looking at just ##np## is not good enough; you also need to look at ##n(1-p)##; that is, you need both successes and failures to have moderate-to-large means.

Anyway, in this particular problem the normal approximation should be quite good. However, you have not shown any substantive calculations, so it is impossible to judge whether you know what to do next, and that is by far the most important issue.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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