Question

Assume that the probability of a being born with Genetic Condition B is π=π=41/60. A study looks at a random sample of 697 volunteers.

Find the most likely number of the 697 volunteers to have Genetic Condition B.
(Round answer to one decimal place.)
μ =

Let XX represent the number of volunteers (out of 697) who have Genetic Condition B. Find the standard deviation for the probability distribution of XX.
(Round answer to two decimal places.)
σ =

Use the range rule of thumb to find the minimum usual value μ-2σ and the maximum usual value μ+2σ.
Enter answer as an interval using square-brackets only with whole numbers.
usual values =

Answer 1

Solution :

Given that,

Using binomial distribution,

Mean = = n * pi = 697 * 41/60 = 476.3

Standard deviation = = n * pi * (1 - pi) = 697 * 41/60 * 19/60 = 12.28

the minimum usual value = - 2 = 476.3 - 2.*12.28 = 452

the maximum usual value = + 2 = 476.3 + 2 * 12.28 = 501