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 71.5 and standard deviation 2.7.

a) If a specimen is acceptable only if its hardness is between 67 and 75, what is the probability that randomly chosen specimen has an acceptable hardness ?

b)If the acceptable range of hardness (71.5 - c, 71.5 + c) for what value of c would 95% of all specimens have an acceptable hardness ?

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 ?

d)What is the probability that at most eight of ten of independently selected specimens have a hardness of less than 73.84 ? Hint: introduce Y = # of specimens with such hardness.

Solution

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    67      
x2 = upper bound =    75      
u = mean =    71.5      
          
s = standard deviation =    2.7      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.666666667      
z2 = upper z score = (x2 - u) / s =    1.296296296      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.047790352      
P(z < z2) =    0.902563288      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.854772936   [ANSWER]

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b)

Here, as z = 1.96 for 95% confidence,

c = z*s = 1.96*2.7 = 5.292 [ANSWER]

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c)

We expect

0.854772936*10 = 8.54772936 acceptable specimens [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 =    73.84      
u = mean =    71.5      
          
s = standard deviation =    2.7      
          
Thus,          
          
z = (x - u) / s =    0.866666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   0.866666667   ) =    0.806937663

Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    10      
p = the probability of a success =    0.806937663      
x = the maximum number of successes =    8      
          
Then the cumulative probability is          
          
P(at most   8   ) =    0.602878239 [ANSWER]