Question

A grinding wheel manufacturer designed a new grinding wheel. Repeated tests were conducted on wheels of approximately the same weight. The tests showed that the new wheel enables free-cutting steels to be cut on an average of 225 surface feet per minute (SFM) with a standard deviation of 16.5 SFM and that the cutting rates are approximately normally distributed.a) What is the 75th percentile of the distribution of cutting rates?---I got 236.137 SFMb) What is the probability that at least 3 wheels out of 10 randomly selected wheels in the study will have a cutting rate that is greater than the cutting rate calculated in part (a)?c) What is the probability that a randomly selected sample of 5 wheels in the study will have a mean cutting rate of at least 225 SFM?

Answer 1

Answer:

a) $a=225 +0.674*16.5=236.121$

So the value of height that separates the bottom 75% of data from the top 25% is 236.121.  

b) $P(X \geq 3) = 1-P(X$

c) $P(\bar X \geq 225)=1- P(\bar X$

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".  

2) Part a

Let X the random variable that represent the cuts of a population, and for this case we know the distribution for X is given by:

$X \sim N(225,16.5)$  

Where $\mu=225$ and $\sigma=16.5$

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

$P(X>a)=0.25$   (a)

$P(X$   (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 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

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:

$z=0.674=\frac{a-225}{16.5}$

And if we solve for a we got

$a=225 +0.674*16.5=236.121$

So the value of height that separates the bottom 75% of data from the top 25% is 236.121.  

Part b

For this case we know that the individual probability of select one wheel with a cutting rate higher than the calculated value in part a is 0.25, and we select n =10 so then we can use the binomial distribution for this case:

$X\sim Bin(n=10, p=0.25)$

And we want this probability:

$P(X \geq 3) = 1-P(X$

We can find the individual probabilities like this:

$P(X=0)=(10C0)(0.25)^0 (1-0.25)^{10-0}=0.0563$

$P(X=1)=(10C1)(0.25)^1 (1-0.25)^{10-1}=0.1877$

$P(X=2)=(10C2)(0.25)^2 (1-0.25)^{10-2}=0.2816$

$P(X \geq 3) = 1-P(X$

Part c

For this case we know that the distribution for the sample mean is given by:

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

And we want this probability:

$P(\bar X \geq 225)$

And for this case we can use the complement rule and the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we replace we got:

$P(\bar X \geq 225)=1- P(\bar X$