Question

question25

Assume that the probability of a being born with Genetic
Condition B is p=p=7/20. A study looks at a random sample of 875
volunteers.

Find the most likely number of the 875 volunteers to have Genetic
Condition B.

*(Round answer to one decimal place.)*

μ =

Let XX represent the number of volunteers (out of 875) 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

p= 7/20 = 0.35

n = 875

formula for Mean ( μ ) is

μ = n * p

μ = 875 * 0.35

μ = 306.25

round answer to one decimal place.

**μ =
306.3**

formula for standard deviation is

σ = 14.10895

round answer to two decimal places.

**σ =
14.11**

using range rule of thumb to find the minimum usual value μ–2σ and the maximum usual value μ+2σ

μ–2σ = 306.3 - ( 2 * 14.11 ) = 278.08

μ+2σ = 306.3 + ( 2 * 14.11 ) = 334.52

round to whole number

**usual values = [
278 , 335 ]**

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