Question

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

Answer:

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Answer 2

(i)The probability that exactly five of them did not see a doctor is 0.15708.

(ii) The probability that fewer than four of them did not see a doctor is 0.432488.

(iii) The expected number of people who would not see a doctor is 3.

Given:

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.

To find;

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?

Solution:

Because of this, 99% of individuals who fracture or dislocate a bone seek medical attention.

There are just two possibilities: either the individual who has a bone fracture or dislocation will go to the doctor, or they won't.

According to prior research, the likelihood of visiting a doctor if one individual has a bone fracture or dislocation is 0.99. Bernoulli's population is the total number of persons who have broken or dislocated a bone, assuming that each person's chance is the same.

Let p indicate the likelihood that a patient will succeed based on the likelihood that they will not seek medical attention after suffering a bone fracture or dislocation.

So, p=1-0.99=0.01

According to Bernoulli's theorem, the probability of exactly r success among the total of n randomly selected from Bernoulli's population is

P(r) = $(^{n} _{p} )p^{r}(1-p)^{n-r}$...... (i)

(i) The total number of persons randomly selected, n=400.

The probability that exactly 5 of them did not see a doctor

So, r=5 , p=0.01

Using equation (i),

P(r-5)= $(^{400} _{5} )(0.01)^{5} (1-0.01)^{400 - 5}$

= [400! / (400-5)! * 5!] $(0.01)^{5} (0.99)^{395}$

=0.15708

(ii) The probability that fewer than four of them did not see a doctor

= P(r <4)

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

= $^{400} _{0} (0.01)^{0} (0.99)^{400}$ + $^{400} _{1} (0.01)^{1} (0.99)^{399}$ + $^{400} _{2} (0.01)^{2 (0.99)^{398}$ + $^{400} _{3} (0.01)^3 (0.99)^{397}$

= 0.017951 + 0.072527 + 0.146154 + 0.195856

= 0.432488

(iii) The expected number of people who would not see a doctor

= np

= 300 * 0.01

=3

Hence, the answer is (i) 0.15708

(ii) 0.432488

(iii) 3

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