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?

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.