The Rockwell hardness of a metal is determined by impressing a hardened point into the surface of the metal and then measuring the depth of penetration of the point. Suppose the Rockwell hardness of a particular alloy is normally distributed with mean 71 and standard deviation 3. (Rockwell hardness is measured on a continuous scale.)

(a) If a specimen is acceptable only if its hardness is between 64 and 78, what is the probability that a randomly chosen specimen has an acceptable hardness? (Round your answer to four decimal places.)

(b) If the acceptable range of hardness is (71 ? c, 71 + c), for what value of c would 95% of all specimens have acceptable hardness? (Round your answer to two decimal places.)

(c) If the acceptable range is as in part (a) and the hardness of each of ten randomly selected specimens is independently determined, what is the expected number of acceptable specimens among the ten? (Round your answer to two decimal places.) specimens

(d) What is the probability that at most eight of ten independently selected specimens have a hardness of less than 73.52? [Hint: Y = the number among the ten specimens with hardness less than 73.52 is a binomial variable; what is p?] (Round your answer to four decimal places.)

a)

b)

And solving for c we got:

So then the value of c = 5.88

c) Let X the niumber of acceptable specimens among a sampel of 10. The acceptable range from part a) is between 64 and 78 and we found that that would represent our p value, and we select samples of size n =10, so then the expected value would be:

d)

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".  

Part a

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

Where and

We are interested on this probability

And the best way to solve this problem is using the normal standard distribution and the z score given by:

If we apply this formula to our probability we got this:

And we can find this probability with this difference:

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

Part b

For this case we want to find the value of c who satisfy this condition:

For this case we can use the z score formula, we can find two values on the normal dtandard distribution that accumulates 0.95 of the values on the middle and for this case the two values are:

So using the z score formula we have this:

And solving for c we got:

And solving for c we got:

So then the value of c = 5.88

Part c

Let X the niumber of acceptable specimens among a sampel of 10. The acceptable range from part a) is between 64 and 78 and we found that that would represent our p value, and we select samples of size n =10, so then the expected value would be:

Part d

For this case we define the random variable Y= the number among the ten specimens with hardness less than 73.52, for this case we can find the probability like this:

[te]x P(X<73.52) = P(Z< \frac{73.52-71}{3}=P(Z<0.84)=0.800" alt="The reason is because [tex] P(-1.96" />

So using the z score formula we have this:

And solving for c we got:

And solving for c we got:

So then the value of c = 5.88

Part c

Let X the niumber of acceptable specimens among a sampel of 10. The acceptable range from part a) is between 64 and 78 and we found that that would represent our p value, and we select samples of size n =10, so then the expected value would be:

Part d

For this case we define the random variable Y= the number among the ten specimens with hardness less than 73.52, for this case we can find the probability like this:

[te]x P(X<73.52) = P(Z< \frac{73.52-71}{3}=P(Z<0.84)=0.800" />

So then we can define our random variable like this:

And we want to find this probability:

The probability mass function for the Binomial distribution is given as:

For this case we can use the complement rule:

idk man good luck man

yes, gavin is right.

step-by-step explanation:

gavin is correct. 

this problem is about converting 9% per year to how many percent per quarter. the rate is still the same. it only differed in the time period. so, through dimensional analysis,

9% per year  × (1 year / 4 quarters) = 2.25% per quarter

if he was asked to convert 9% per year to how many percent per quarter  compounded quarterly, then you will not simply divide by four. this is because compounding is not linear.

hope this , have a good day. c;