Question

5.26 A high percentage of people who fracture or dislocate a bone see a doctor for that condition. Suppose the percentage is 99%. Consider a sample in which 300 people are randomly selected who have fractured or dislocated a bone. a. What is the probability that exactly five of them did not see a doctor? b. What is the probability that fewer than four of them did not see a doctor? c. What is the expected number of people who would not see a doctor?

Answer 1

Solution:

Let X be a random variable who have fractured or dislocated a bone did not see a doctor.

P1: Percentage of people who fracture or dislocate a bone see a doctor for that condition=0.99

p: Percentage of people who fracture or dislocate a bone did not see a doctor for that condition=0.01

n=Number of people who fracture or dislocate a bone=300

First, we have to check that the information approximates to normal distribution or not.

If n*p>5 and n*(1-p)>5 then Normal approximation to Binomial.

Here, n*p=3<5 and n*(1-p)>5 so, the condition is not satisfied. So we use the Binomial distribution while solving this problem.

Here, X~B(n,p)

a) P(exactly five of them did not see a doctor)=P(X=5)

  

b)P( fewer than four of them did not see a doctor)=P(X<4)

= P(X=0)+P(X=1)+P(X=2)+P(X=3)

=0.0490+0.1486+0.2244+0.2251

=0.6472

c) Expected number of people who would not see a doctor=Mean of binomial distribution

=n*p   

=300*0.01

=3