Currently quarters have weights that are normally distributed with a mean of 5.6

Question

Currently quarters have weights that are normally distributed with a mean of 5.670 and a standard deviation of 0.062 g. A vending machine is configured to accept only those quarters with weights between 5.550 g and 5.790 g. a) if 280 different quarters are inserted into the vending machine what is the expected number of rejected quarters? b) if 280 different quarters are inserted into the vending machine what is the probability that the mean falls between the limits of 5.550 and 5.790 g? c) if you own the vending machine which result would concern you more? the result from part a or part b and why?

Explanation / Answer

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    5.55      
x2 = upper bound =    5.79      
u = mean =    5.67      
          
s = standard deviation =    0.062      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.935483871      
z2 = upper z score = (x2 - u) / s =    1.935483871      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.026465473      
P(z < z2) =    0.973534527      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.947069054      

Thus, those outside this interval is the complement =    0.052930946

Hence, we expect 0.052930946*280 = 14.82 are epected to be rejected.

ANSWER: 14.82

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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 =    5.55      
x2 = upper bound =    5.79      
u = mean =    5.67      
n = sample size =    280      
s = standard deviation =    0.062      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -32.38683974      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    32.38683974      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    2.1027E-230      
P(z < z2) =    1      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    1   [ANSWER]

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c)
  
PART B WILL CONCERN ME MORE, as it has a lower probability of happening.

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