About 30% of all patients admitted to the JJ Medical Clinic (JJMC) fail to pay t

Question

About 30% of all patients admitted to the JJ Medical Clinic (JJMC) fail to pay their bills(default). Whether any patient pays is independent of whether any other patient pays. Eventually these defaulted bills are forgiven. Every bill is $500 (defaulted or not.) Suppose that on a certain day, 50 patients are admitted to the JJ Medical Clinic. Let X equal the number of these patients who will default on their bill.

What is the expected value of X?  
b. What is the standard deviation of X?  
c. What is the probability that exactly 16 patients default?
d. What is the probability that no more than 16 patients default?  
e. What is the probability that fewer than 10 patients default?   
f. What is the probability that between 10 and 15 patients default?  
g. In terms of X, give a formula for the total amount the hospital takes in.  
h. What is the expected amount that the hospital takes in?  
i. What is the probability that the clinic takes in more than $19,000?

Explanation / Answer

Here, n = 50, p = 0.30.

a)

Hence,

E(x) = n p = 50*0.30 = 15 [ANSWER]

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b)

s = standard deviation = sqrt(np(1-p)) =    3.240370349 [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 =    15.5      
x2 = upper bound =    16.5      
u = mean = np =    15      
          
s = standard deviation = sqrt(np(1-p)) =    3.240370349      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.15430335      
z2 = upper z score = (x2 - u) / s =    0.46291005      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.561314722      
P(z < z2) =    0.678285578      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.116970856   [ANSWER]

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d)

We first get the z score for the critical value:          
          
x = critical value =    16.5      
u = mean = np =    15      
          
s = standard deviation = sqrt(np(1-p)) =    3.240370349      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    0.46291005      
          
Thus, the left tailed area is          
          
P(z <   0.46291005   ) =    0.678285578 [ANSWER]

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e)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    9.5      
x2 = upper bound =    15.5      
u = mean = np =    15      
          
s = standard deviation = sqrt(np(1-p)) =    3.240370349      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.69733685      
z2 = upper z score = (x2 - u) / s =    0.15430335      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.044816497      
P(z < z2) =    0.561314722      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.516498225   [ANSWER]  

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