The Rockwell hardness of a metal is determined by impressing

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 69 and standard deviation 3. (Rockwell hardness is measured on a continuous scale.) If a specimen is acceptable only if its hardness is between 66 and 72, what is the probability that a randomly chosen specimen has an acceptable hardness? (Round your answer to four decimal places.) If the acceptable range of hardness is (69 - c, 69 + c), for what value of c would 95% of all specimens have acceptable hardness? (Round your answer to two decimal places.) 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.) What is the probability that at most eight of ten independently selected specimens have a hardness of less than 72.84?

Solution

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    62      
x2 = upper bound =    74      
u = mean =    68      
          
s = standard deviation =    3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2      
z2 = upper z score = (x2 - u) / s =    2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022750132      
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.954499736   [ANSWER]

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b)

For the middle 95%, the corresponding z value is, by table/technology,

z = 1.959963985

Thus,

c = z*sigma = 1.959963985*3 = 5.879891954 [ANSWER]

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c)

E(x) = n p = 10*0.954499736 = 9.54499736 [ANSWER]

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d)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    70.52      
u = mean =    68      
          
s = standard deviation =    3      
          
Thus,          
          
z = (x - u) / s =    0.84      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   0.84   ) =    0.799545807

Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    10      
p = the probability of a success =    0.799545807      
x = the maximum number of successes =    8      
          
Then the cumulative probability is          
          
P(at most   8   ) =    0.62556042 [ANSWER]