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The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3 mm and standard deviation 0.1 mm. A random sample of 200 wafers is drawn. What is the probability that the sample mean warpage exceeds 1.305 mm?

asked
User JayTaph
by
7.9k points

1 Answer

4 votes

Answer:

0.2389

Explanation:

The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3 mm and standard deviation 0.1 mm.


\mu=1.3mm\\\sigma=0.1mm

A random sample of 200 wafers is drawn

we are supposed to find What is the probability that the sample mean warpage exceeds 1.305 mm

We will use central limit theorem

According to central limit theorem:


\sigma^2_{\bar{x}}=(\sigma^2)/(n)


\sigma^2_{\bar{x}}=(0.1^2)/(200)


\sigma_{\bar{x}=\sqrt{(0.1^2)/(200)}


\sigma_{\bar{x}=√(0.00005)

Now we are supposed to find What is the probability that the sample mean warpage exceeds 1.305 mm i.e.P(x>1.305)


z=(x-\mu)/(\sigma)


z=(1.305-1.3)/(√(0.00005))


z=0.71

Refer the z table

P(z<0.71)=0.7611


P(\bar{x}>1.305)=1-P(z<0.71) =1-0.7611=0.2389

Hence the probability that the sample mean warpage exceeds 1.305 mm is 0.2389

answered
User Ashwin Kumar
by
8.5k points
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