The breaking strength of a rivet has a mean value of 10 000

The breaking strength of a rivet has a mean value of 10. 000 psi and a standard deviation of 500 psi. Which probability distribution will you suggest to model the average strength of 100 randomly selected rivets, and why? What is the (approximate) probability that the sample mean is between 9,950 and 10,150?

Solution

a)

A normal distribution, as by central limit theorem, the sampling distirbution of the means is approximately normally distirbuted, especially as the sample size becomes larger.

b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    9950      
x2 = upper bound =    10150      
u = mean =    10000      
n = sample size =    100      
s = standard deviation =    500      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    3      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.998650102      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.839994848   [ANSWER]