Do these problems by hand. For each probability from the Normal distribution, dr

Question

Do these problems by hand. For each probability from the Normal distribution, draw a Normal curve and shade the region representing probability of interest. Do not use the same Normal curve for multiple probabilities, only one probability should be represented on each plot.

Extruded plastic rods are automatically cut, with a target length of 6 inches. The actual lengths are normally distributed about a mean of 6 inches, and their standard deviation is 0.06 inches. The manufacturing tolerance limits are 5.9 to 6.1 inches.

[a.] What proportion of the plastic rods have lengths outside of the tolerance limits?

[b.] What does the standard deviation need to be reduced to for 99% of the plastic rods to be within the tolerance limits?

Explanation / Answer

A)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    5.9      
x2 = upper bound =    6.1      
u = mean =    6      
          
s = standard deviation =    0.06      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.666666667      
z2 = upper z score = (x2 - u) / s =    1.666666667      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.047790352      
P(z < z2) =    0.952209648      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.904419295      

Thus, those outside this interval is the complement =    0.095580705 = 9.558% [ANSWER]

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b)

The z score in that case for the middle 99% is

z = 2.575829304

Hence, for x = 6.1,

sigma = (x-u)/z = (6.1-6)/2.575829304 = 0.038822448 in [ANSWER]
      

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