SOLUTION: Assume that the probability of a being born with Genetic Condition B is π=7/60
A study looks at a random sample of 719 volunteers.
Find the most likely number of the 719
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Question 1105043: Assume that the probability of a being born with Genetic Condition B is π=7/60
A study looks at a random sample of 719 volunteers.
Find the most likely number of the 719 volunteers to have Genetic Condition B.
μ =
Let X represent the number of volunteers (out of 719) who have Genetic Condition B. Find the standard deviation for the probability distribution of X
(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. ? Round your answer to one decimal place. usual values =
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
mean is (7/60) * 719 = 83.8833 approximately 84 have genetic condition B
:
proportion is 7/60 is approximately 0.12
:
we use the binomial distribution's standard deviation
:
standard deviation is sqrt(719 * 0.12 * (1-0.12)) = 8.7136 approximately 8.71
:
2 * standard deviation is 17.42
:
min/max usual value is [67, 101 ]
:
min/max usual value rounded (66.6, 101.4)
:
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