What is the age distribution of patients who make office visits to a doctor or n

Question

What is the age distribution of patients who make office visits to a doctor or nurse? The following table is based on information taken from a medical journal.

Suppose you are a district manager of a health management organization (HMO) that is monitoring the office of a local doctor or nurse in general family practice. This morning the office you are monitoring has eight office visits on the schedule. What is the probability of the following?(a) At least half the patients are under 15 years old. (Round your answer to three decimal places.)
1

Explain how this can be modeled as a binomial distribution with 8 trials, where success is visitor age is under 15 years old and the probability of success is 15%?

Let n = 8, p = 0.85 and compute the probabilities using the binomial distribution. Let n = 8, p = 0.15 and compute the probabilities using the binomial distribution.     Let n = 8, p = 0.20 and compute the probabilities using the binomial distribution. Let n = 15, p = 0.15 and compute the probabilities using the binomial distribution.

(b) From 2 to 5 patients are 65 years old or older (include 2 and 5). (Round your answer to three decimal places.)
3

(c) From 2 to 5 patients are 45 years old or older (include 2 and 5). (Hint: Success if 45 or older. Use the table to compute the probability of success on a single trial. Round your answer to three decimal places.)
4

(d) All the patients are under 25 years of age. (Round your answer to three decimal places.)
5

(e) All the patients are 15 years old or older. (Round your answer to three decimal places.)
6

In the book A Guide to the Development and Use of the Myers-Briggs Type Indicators by Myers and McCaully, it was reported that approximately 45% of all university professors are extroverted. Suppose you have classes with six different professors.

(a) What is the probability that all six are extroverts? (Round your answer to three decimal places.)
1

(b) What is the probability that none of your professors is an extrovert? (Round your answer to three decimal places.)
2

(c) What is the probability that at least two of your professors are extroverts? (Round your answer to three decimal places.)
3

(d) In a group of six professors selected at random, what is the expected number of extroverts?
4 extroverts

What is the standard deviation of the distribution? (Round your answer to two decimal places.)
5 extroverts

Explanation / Answer

A)

That means at least 4 are under 15.

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    8      
p = the probability of a success =    0.15      
x = our critical value of successes =    4      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   3   ) =    0.97864753
          
Thus, the probability of at least   4   successes is  
          
P(at least   4   ) =    0.02135247 [ANSWER]

********************

Let n = 8, p = 0.15 and compute the probabilities using the binomial distribution.

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b)

Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)          
          
Here,          
          
x1 =    2      
x2 =    5      
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    8      
p = the probability of a success =    0.45      
          
Then          
          
P(at most    1   ) =    0.063181062
P(at most    5   ) =    0.911544137
          
Thus,          
          
P(between x1 and x2) =    0.848363074   [ANSWER]

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c)

Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)          
          
Here,          
          
x1 =    2      
x2 =    5      
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    8      
p = the probability of a success = 0.15+0.45 =   0.6      
          
Then          
          
P(at most    1   ) =    0.00851968
P(at most    5   ) =    0.68460544
          
Thus,          
          
P(between x1 and x2) =    0.67608576   [ANSWER]  
  
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d)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    8      
p = the probability of a success = 0.15+0.05 =   0.2      
x = the number of successes =    8      
          
Thus, the probability is          
          
P (    8   ) =    0.00000256
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e)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    8      
p = the probability of a success = 1 - 0.15 =   0.85      
x = the number of successes =    8      
          
Thus, the probability is          
          
P (    8   ) =    0.272490525 [ANSWER]

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