In the following problem, check that it is appropriate to use the normal approxi

Question

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 42% 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 130 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 (b) fewer than 45 of the claims have been padded (c) from 40 to 64 of the claims have been padded (d) more than 80 of the claims have not been padded

Explanation / Answer

a)

Here, n = 130, p = 0.42.

This means we need at least 65 padded claims.

We first get the z score for the critical value:          
          
x = critical value =    64.5      
u = mean = np =    54.6      
          
s = standard deviation = sqrt(np(1-p)) =    5.627432807      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    1.759239131      
          
Thus, the right tailed area is          
          
P(z >   1.759239131   ) =    0.03926845 [ANSWER]

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b)

We first get the z score for the critical value:          
          
x = critical value =    44.5      
u = mean = np =    54.6      
          
s = standard deviation = sqrt(np(1-p)) =    5.627432807      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -1.794779315      
          
Thus, the left tailed area is          
          
P(z <   -1.794779315   ) =    0.036344434 [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 =    54.6      
          
s = standard deviation = sqrt(np(1-p)) =    5.627432807      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.683283927      
z2 = upper z score = (x2 - u) / s =    1.759239131      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.003645153      
P(z < z2) =    0.96073155      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.957086396   [ANSWER]

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d)

We first get the z score for the critical value:          
          
x = critical value =    80.5      
u = mean = np =    54.6      
          
s = standard deviation = sqrt(np(1-p)) =    5.627432807      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    4.602453887      
          
Thus, the left tailed area is          
          
P(z <   4.602453887   ) =    2.08771*10^-6 [ANSWER]
  

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