Question

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.

a)If a specimen is acceptable only if its hardness is between 70 and 74, 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.)
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.)

Answer 1

a)

µ =    71                              
σ =    3                              
we need to calculate probability for ,                                  
P (   70   < X <   74   )                  
=P( (70-71)/3 < (X-µ)/σ < (74-71)/3 )                                  
                                  
P (    -0.333   < Z <    1.000   )                   
= P ( Z <    1.000   ) - P ( Z <   -0.333   ) =    0.8413   -    0.3694   =    0.4719

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b)

µ=   71                  
σ =    3                  
proportion=   0.025                  
                      
Z value at    0.025   =   -1.96   (excel formula =NORMSINV(   0.025   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -1.96   *   3   +   71  
X   =   65.12

µ=   71                  
σ =    3                  
proportion=   0.975                  
                      
Z value at    0.975   =   1.96   (excel formula =NORMSINV(   0.975   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.96   *   3   +   71  
X   =   76.88  

Answer will be [ 65.12 ,76.88]

c = 5.88
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c)

Expected number = 10 * .4719 = 4.719 = 4.72

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d)

µ =    71      
σ =    3      
          
P( X ≤    73.52   ) = P( (X-µ)/σ ≤ (73.52-71) /3)  
=P(Z ≤   0.840   ) =   0.7995458

Now p = 0.7995458

n =10

P(X=x) = C(n,x)*px*(1-p)(n-x)      

X	P(X)
9	0.2677
10	0.1068

P(at most 8 ) = P(x<=8) = 1 - P(x=9) - P(x=10)

P(x<=8) = 1- 0.2677 -0.1068
= 0.6256

Please revert back in case of any doubt.

Please upvote. Thanks in advance.