Question

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.
a. What is the probability that the sample mean warpage exceeds 1.305 mm?

b. Find the 25th percentile of the sample mean.

c. How many wafers must be sampled so that the probability is 0.05 that the sample mean exceeds 1.305?

Answer 1

Answer:

a) $P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)$

And using the complement rule, a calculator, excel or the normal standard table we have that:

$P(Z>0.707)=1-P(Z$

b) $z=-0.674$

And if we solve for a we got

$a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=$

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

c) $P( \bar X >1.305) = 0.05$  

We can use the z score formula:

$P( \bar X >1.305) = 1-P(\bar X$

Then we have this:

$P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95$

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

$n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

Let X the random variable that represent the amount of warpage of a population and we know

Where $\mu=1.3$ and $\sigma=0.1$

Since the sample size is large enough we can use the central limit theorem andwe know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We can find the probability required like this:

$P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)$

And using the complement rule, a calculator, excel or the normal standard table we have that:

$P(Z>0.707)=1-P(Z$

Part b

For this part we want to find a value a, such that we satisfy this condition:

$P(\bar X>a)=0.75$   (a)

$P(\bar X$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

$P(X$  

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.674$

And if we solve for a we got

$a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=$

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

Part c

For this case we want this condition:

$P( \bar X >1.305) = 0.05$  

We can use the z score formula:

$P( \bar X >1.305) = 1-P(\bar X$

Then we have this:

$P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95$

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

$n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076$