ball bearings are manufactured with a mean diameter of 5 mm

ball bearings are manufactured with a mean diameter of 5 mm because of variablility in the manufacturing process, the diameters of the ball bearing are approximately normally distributed with a standard deviation of 0.025mm

what proportion of call bearings have a diameter more then 5.013 mm? round 3 places

all ball bearings that have a diameter less then 4.94 mm or greater then 5.06 mm are discarded. what proportion of ball bearings will be discarded? round 4 places

using the rounded results of the question above if 30000 ball bearings are manufactured in a day how many should the plant expect to discard in a day? round to whole number

what diameters seperate the middle 95% of the ball bearings where middle means those ball bearings with diameter closest to the desired 5 mm round 3 places

the low diameter seperating middle 95%

the high diameter seperating middle 95%

Solution

a)

what proportion of call bearings have a diameter more then 5.013 mm? round 3 places

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    5.013      
u = mean =    5      
          
s = standard deviation =    0.025      
          
Thus,          
          
z = (x - u) / s =    0.52      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.52   ) =    0.301531788 [ANSWER]

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b)

all ball bearings that have a diameter less then 4.94 mm or greater then 5.06 mm are discarded. what proportion of ball bearings will be discarded? round 4 places

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    4.94      
x2 = upper bound =    5.06      
u = mean =    5      
          
s = standard deviation =    0.025      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.4      
z2 = upper z score = (x2 - u) / s =    2.4      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.008197536      
P(z < z2) =    0.991802464      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.983604928      

Thus, those outside this interval is the complement = 0.016395072   [ANSWER]

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c)  

using the rounded results of the question above if 30000 ball bearings are manufactured in a day how many should the plant expect to discard in a day? round to whole number

There are 0.016395072*30000 = 491.85216 = 492 rejects. [ANSWER, 492]

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what diameters seperate the middle 95% of the ball bearings where middle means those ball bearings with diameter closest to the desired 5 mm round 3 places

The middle 95% are bounded by scores with left tailed areas of 0.025 and 0.975.

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.025      
          
Then, using table or technology,          
          
z =    -1.959963985      
          
As x = u + z * s,          
          
where          
          
u = mean =    5      
z = the critical z score =    -1.959963985      
s = standard deviation =    0.025      
          
Then          
          
x = critical value =    4.9510009   [ANSWER, LOW DIAMETER]

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First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.975      
          
Then, using table or technology,          
          
z =    1.959963985      
          
As x = u + z * s,          
          
where          
          
u = mean =    5      
z = the critical z score =    1.959963985      
s = standard deviation =    0.025      
          
Then          
          
x = critical value =    5.0489991   [ANSWER, HIGH DIAMETER]