The diameter of a dot created by a pen mark is normally dist

The diameter of a dot created by a pen mark is normally distributed with a mean diameter of 0.002 inch and a standard deviation of 0.0004. What is the probability that the diameter of a dot exceeds 0.0026 inch? What is the probability that a diameter is between 0.0014 and 0.0026 inch? What standard deviation of diameters is needed so that the probability in part b is 0995?

Solution

A)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    0.0026      
u = mean =    0.002      
          
s = standard deviation =    0.0004      
          
Thus,          
          
z = (x - u) / s =    1.5      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.5   ) =    0.066807201 [ANSWER]

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B)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    0.0014      
x2 = upper bound =    0.0026      
u = mean =    0.002      
          
s = standard deviation =    0.0004      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.5      
z2 = upper z score = (x2 - u) / s =    1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.066807201      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.866385597   [ANSWER]

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c)

0.0014 and 0.0026 are symmetric about the mean.

Now, if the middle area is 0.995, then the left tailed area of 0.0014 will be 0.0025.

Thus, its z score is

z(0.0014) = -2.807033768

Thus, as

sigma = (x - u) / z = (0.0014 - 0.002)/-2.807033768 = 0.000213749 [ANSWER]