Question

Medical researcher estimates that .00004 of the population has a rare blood disorder. if the researcher randomly selects 100,000 people from the population? appendix a statistical tables (round your answers to 4 decimal places.)

a. what is the probability that seven or more people will have the rare blood disorder?

Answer 1

To solve this problem, we make use of the binomial probability equation.

P = [n! / (n – r)! r!] p^r * q^(n – r)

where,

n is the total number of sample = 100,000

r is the number of people with rare blood disorder = 7 or more

p is success of having a disorder = 0.00004

q is 1- p = 0.99996

But since it is easier to solve for the sum of probabilities with 0 to 6 disorder then deduct it from 1.

So,

=> at r = 0

P (r = 0) = [100,000! / (100,000 – 0)! 0!] (0.00004)^0 * (0.99996)^(100,000 – 0)

P (r = 0) = 0.01831

=> at r = 1

P (r = 1) = [100,000! / (100,000 – 1)! 1!] (0.00004)^1 * (0.99996)^(100,000 – 1)

P (r = 1) = 0.07326

=> at r = 2

P (r = 2) = [100,000! / (100,000 – 2)! 2!] (0.00004)^2 * (0.99996)^(100,000 – 2)

P (r = 2) = 0.14652

=> at r = 3

P (r = 3) = [100,000! / (100,000 – 3)! 3!] (0.00004)^3 * (0.99996)^(100,000 – 3)

P (r = 3) = 0.19537

=> at r = 4

P (r = 4) = [100,000! / (100,000 – 4)! 4!] (0.00004)^4 * (0.99996)^(100,000 – 4)

P (r = 4) = 0.19537

=> at r = 5

P (r = 5) = [100,000! / (100,000 – 5)! 5!] (0.00004)^5 * (0.99996)^(100,000 – 5)

P (r = 5) = 0.15630

=> at r = 6

P (r = 6) = [100,000! / (100,000 – 6)! 6!] (0.00004)^6 * (0.99996)^(100,000 – 6)

P (r = 6) = 0.10420

So the total P is:

P (r = 0 to 6) = 0.01831 + 0.07326 + 0.14652 + 0.19537 + 0.19537 + 0.15630 + 0.10420

P (r = 0 to 6) = 0.88933

=> So the probability that 7 or more people will have the disease is:

P (r ≥ 7) = 1 - 0.88933

P (r ≥ 7) = 0.11067 = 11.067%

There is only 0.11067 or 11.067% probability that 7 or more will have the disorder.