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:

a) [tex]a=225 +0.674*16.5=236.121[/tex]

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

b) [tex] P(X \geq 3) = 1-P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)]= 1-0.5256=0.4744[/tex]

c) [tex]P(\bar X \geq 225)=1- P(\bar X <225) = 1-P(Z<\frac{225-225}{\frac{16.5}{\sqrt{5}}}) = 1-P(Z<0) = 1-0.5 = 0.5[/tex]

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:

[tex]X \sim N(225,16.5)[/tex]  

Where [tex]\mu=225[/tex] and [tex]\sigma=16.5[/tex]

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

[tex]P(X>a)=0.25[/tex]   (a)

[tex]P(X<a)=0.75[/tex]   (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:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.75[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.75[/tex]

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

[tex]z=0.674=\frac{a-225}{16.5}[/tex]

And if we solve for a we got

[tex]a=225 +0.674*16.5=236.121[/tex]

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:

[tex] X\sim Bin(n=10, p=0.25)[/tex]

And we want this probability:

[tex] P(X \geq 3) = 1-P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]

We can find the individual probabilities like this:

[tex]P(X=0)=(10C0)(0.25)^0 (1-0.25)^{10-0}=0.0563[/tex]

[tex]P(X=1)=(10C1)(0.25)^1 (1-0.25)^{10-1}=0.1877[/tex]

[tex]P(X=2)=(10C2)(0.25)^2 (1-0.25)^{10-2}=0.2816[/tex]

[tex] P(X \geq 3) = 1-P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)]= 1-0.5256=0.4744[/tex]

Part c

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

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

And we want this probability:

[tex]P(\bar X \geq 225)[/tex]

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

[tex] z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And if we replace we got:

[tex]P(\bar X \geq 225)=1- P(\bar X <225) = 1-P(Z<\frac{225-225}{\frac{16.5}{\sqrt{5}}}) = 1-P(Z<0) = 1-0.5 = 0.5[/tex]

This problem involves using statistics and probability to interpret and make predictions from data involving the distribution of cutting rates of grinding wheels. It involves understanding mean and standard deviation values and the concept of percentiles in the context of normal distribution. To tackle similar problems, you need to understand how to calculate z-scores, use standard normal tables or functions and probabilities involving multiple events, which may involve the use of formulas such as the binomial probability formula.

Please note that the information provided in your question doesn't directly relate to the grinding wheel problems outlined. However, generally, these types of problems involve an understanding of statistics, probability, and the properties of the normal distribution, especially when using mean and standard deviation values to make calculations and predictions. It also involves understanding percentiles of a distribution and how these relate to the standard normal distribution and z-scores.

For example, if we consider that your calculated 75th percentile of the distribution of cutting rates (236.137 SFM) is correct, you can find the probability of a wheel having a cutting rate greater than this value by finding the corresponding z-score and looking this up on a table of standard normal probabilities, or using a computer program or calculator function that will provide this value. To calculate a probability involving multiple randomly selected wheels, we might need to use a binomial probability formula or similar.

Understanding these methods will allow you to generalise to other such problems. Given the complexity of these concepts however, I would recommend finding a tutor or seeking assistance from your teacher or lecturer to guide you through them.

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