In the following problem check that it is appropriate to use

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. Do you try to pad an insurance claim to cover your deductible? About 40% of all U.S. adults will try to pad their insurance claims! Suppose that you are the director of an insurance adjustment office. Your office has just received 134 insurance claims to be processed in the next few days. Find the following probabilities. (Round your answers to four decimal places.) (a) half or more of the claims have been padded Incorrect: Your answer is incorrect. (b) fewer than 45 of the claims have been padded Incorrect: Your answer is incorrect. (c) from 40 to 64 of the claims have been padded (d) more than 80 of the claims have not been padded

Solution

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound = (134/2)-0.5 =   66.5      
x2 = upper bound =    67.5      
u = mean = np =    53.6      
          
s = standard deviation = sqrt(np(1-p)) =    5.670978752      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    2.274739611      
z2 = upper z score = (x2 - u) / s =    2.451076015      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.988539223      
P(z < z2) =    0.992878506      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.004339284   [ANSWER]

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b)

We first get the z score for the critical value:          
          
x = critical value =    44.5      
u = mean = np =    53.6      
          
s = standard deviation = sqrt(np(1-p)) =    5.670978752      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -1.604661276      
          
Thus, the left tailed area is          
          
P(z <   -1.604661276   ) =    0.054284184 [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    39.5      
x2 = upper bound =    64.5      
u = mean = np =    53.6      
          
s = standard deviation = sqrt(np(1-p)) =    5.670978752      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.486343296      
z2 = upper z score = (x2 - u) / s =    1.922066803      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.00645317      
P(z < z2) =    0.972701324      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.966248154   [ANSWER]

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d)

We first get the z score for the critical value:          
          
x = critical value =    80.5      
u = mean = np =    53.6      
          
s = standard deviation = sqrt(np(1-p)) =    5.670978752      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    4.743449267      
          
Thus, the left tailed area is          
          
P(z <   4.743449267   ) =    1.05055*10^-6 [ANSWER]
  

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Note: I used continuity correction for this one. If you do not use continuity correction in your class, please resubmit this question and note \"do not use continuity correction\". That way we can continue helping you! Thanks!