For the normal distribution described in Exercise 712 what t

For the normal distribution described in Exercise 7.12, what tax preparation fee would have been exceeded by 90% of the tax preparation customers?

Solution

Please attach Ex. 7.12 for 7.32, and submit it as another question. Thanks!

I am going to answer 7.33.

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7.33

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    6350      
u = mean =    6050      
          
s = standard deviation =    1500      
          
Thus,          
          
z = (x - u) / s =    0.2      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.2   ) =    0.420740291 [answer]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.99      
          
Then, using table or technology,          
          
z =    2.326347874      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    6050      
z = the critical z score =    2.326347874      
s = standard deviation =    1500      
          
Then          
          
x = critical value = $9539.52 [ANSWER]