Question 1192139
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].