An assembly consists of three components placed side by side

An assembly consists of three components placed side by side. The length of each component is normally distributed with mean 2 inches and standard deviation 0.2 inch. Specifications require that all assemblies are between 5.7 and 6.3 inches long. How many assembles will pass these requirements?

Solution

A.

The mean sum of their lengths must then be

M(tot) = m1 + m2 + m3 = 2 + 2 + 2 = 6 in

And the standard deviation of their sum is

s(tot) = sqrt(s1^2 + s2^2 + s3^2) = sqrt(0.2^2 + 0.2^2 + 0.2^2) = 0.34641 in

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    5.7      
x2 = upper bound =    6.3      
u = mean =    6      
n = sample size =    1 [only one assembly]      
s = standard deviation =    0.34641      
          
Thus, the two z scores are          
          
z1 = lower z score =    -0.866025808      
z2 = upper z score =    0.866025808      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.193238005      
P(z < z2) =    0.806761995      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.61352399 of assemblies, or approx. 61.352% of them. [ANSWER]