Rockwell hardness of pins of a certain type is known to have

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.

a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 9 pins is at least 51?
b) Without assuming population normality, what is the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51?

Solution

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    51      
u = mean =    50      
n = sample size =    9      
s = standard deviation =    1.2      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.5      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.5   ) =    0.006209665 [ANSWER]

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b)

BY central limit theorem, the sampling distirbution of means is still approximately normal even without assuming anything about the original distribution.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    51      
u = mean =    50      
n = sample size =    40      
s = standard deviation =    1.2      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    5.270462767      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   5.270462767   ) =    6.80401*10^-8 [ANSWER]