Twenty percent of all adult males did not visit their physicians’ offices last year. The probability that less than 23% of adult males in a sample of 800 did not visit their physicians' offices last year is approximately:

.9838

I have no clue

To find the probability that less than 23% of adult males in a sample of 800 did not visit their physicians' offices last year, we can use a binomial distribution.

First, let's define the variables:
- Let n be the number of trials, which is 800 in this case.
- Let p be the probability of success on each trial, which is 20% or 0.2 (since 20% did not visit their physicians' offices).
- Let X be the number of successes, which is the number of adult males who did not visit their physicians' offices.

Now, we can calculate the probability using the binomial probability formula:

P(X < 23% of 800) = P(X < 0.23 * 800)

To calculate this probability, we need to use the cumulative distribution function (CDF) of the binomial distribution. However, calculating the CDF for large values of n can be computationally intensive. Instead, we can use an approximation called the normal approximation to the binomial distribution.

The conditions for using the normal approximation are:
1. np ≥ 10, which means the expected number of successes is greater than or equal to 10.
2. n(1-p) ≥ 10, which means the expected number of failures is greater than or equal to 10.

In this case, np = 800 * 0.2 = 160 and n(1-p) = 800 * 0.8 = 640, so both conditions are satisfied.

To approximate the probability using the normal distribution, we need to calculate the mean and standard deviation of the binomial distribution.

Mean (μ) = np = 800 * 0.2 = 160
Standard Deviation (σ) = sqrt(np(1-p)) = sqrt(800 * 0.2 * 0.8) = sqrt(128) = 11.3137

Now, we can convert the desired value of X (0.23 * 800) to a z-score using the formula:
z = (X - μ) / σ

z = (0.23 * 800 - 160) / 11.3137

Calculate the value of z and consult a standard normal distribution table or use a statistical software to find the probability associated with that z-score.

For example, if you use a standard normal distribution table, you can look up the z-score value 1.57 and find the corresponding probability, which is approximately 0.9418.

So, the probability that less than 23% of adult males in a sample of 800 did not visit their physicians' offices last year is approximately 0.9418 or 94.18%.