Question 1192139: Assume that the probability of a being born with Genetic Condition B is `pi=`7/20``. A study looks at a random sample of 423 volunteers.
Find the most likely number of the 423 volunteers to have Genetic Condition B.
(Round answer to one decimal place.)
μ =
Let `X` represent the number of volunteers (out of 423) 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.
usual values =
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
**1. Most Likely Number with Genetic Condition B:**
The most likely number of volunteers with Genetic Condition B is simply the expected value (mean) of the distribution. For a binomial distribution (which this is, since each volunteer either has the condition or doesn't), the mean is:
μ = n * p
Where:
* n = sample size = 423
* p = probability of having the condition = 7/20 = 0.35
μ = 423 * 0.35 = 148.05
Rounding to one decimal place, the most likely number is 148.1.
**2. Standard Deviation:**
The standard deviation for a binomial distribution is:
σ = √(n * p * (1 - p))
σ = √(423 * 0.35 * (1 - 0.35))
σ = √(423 * 0.35 * 0.65)
σ = √96.1575
σ ≈ 9.81
Rounding to two decimal places, the standard deviation is 9.81.
**3. Range Rule of Thumb:**
The range rule of thumb states that most values fall within two standard deviations of the mean.
* **Minimum usual value:** μ - 2σ = 148.05 - 2 * 9.81 = 148.05 - 19.62 = 128.43
* **Maximum usual value:** μ + 2σ = 148.05 + 2 * 9.81 = 148.05 + 19.62 = 167.67
Since we're dealing with whole numbers (number of volunteers), we round these to the nearest whole number.
Therefore, the usual values are [128, 168].
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