The lengths of steel beams made in a factory have a normal d

The lengths of steel beams made in a factory have a normal distribution, where their mean is 4.52 meters with a std. dev. of 0.04 meters. Company A orders steel beams for which the lengths are between 4.55 and 4.45 meters.

1. Determine the ratio of the steel beams made in the factory that meet Company A’s requirements

2. Find the 99th percentile for the length of the steel beams

3. If, through new methods, the factory manages to set the mean to 4.5 meters, what ratio of the steel beams will meet Company A’s requirements?

4. With the new mean of 4.5 meters, determine the std. dev. such that 99 percent of the steel beams will meet Company A’s requirements.

Solution

1.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    4.45      
x2 = upper bound =    4.55      
u = mean =    4.52      
          
s = standard deviation =    0.04      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.75      
z2 = upper z score = (x2 - u) / s =    0.75      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.040059157      
P(z < z2) =    0.773372648      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.733313491   [ANSWER]

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2)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.99      
          
Then, using table or technology,          
          
z =    2.326347874      
          
As x = u + z * s,          
          
where          
          
u = mean =    4.52      
z = the critical z score =    2.326347874      
s = standard deviation =    0.04      
          
Then          
          
x = critical value =    4.613053915   [ANSWER]

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3.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    4.45      
x2 = upper bound =    4.55      
u = mean =    4.5      
          
s = standard deviation =    0.04      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.25      
z2 = upper z score = (x2 - u) / s =    1.25      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.105649774      
P(z < z2) =    0.894350226      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.788700453   [ANSWER]

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4.

The z score for this is, by table/technology,

z = 2.575829304

Thus,

sigma = (x - u)/z = (4.55 - 4.5)/2.575829304 = 0.019411224 [ANSWER]