Question

43% of adults did not visit their physicians offices last year. let x be the number of adults in a random sample of 44 adults who did not visit their physicians offices last year. the mean and standard deviation of the probability distribution of x, rounded to two decimal places, are: the standard deviation is blank, the mean is 18.92

Explanation:

Step1: Identify Distribution Type

This is a binomial distribution problem where \( n = 44 \) (sample size) and \( p = 0.43 \) (probability of an adult not visiting a physician). For a binomial distribution, the mean \( \mu = np \) and standard deviation \( \sigma = \sqrt{np(1 - p)} \).

Step2: Calculate Standard Deviation

First, find \( 1 - p = 1 - 0.43 = 0.57 \). Then, compute \( np(1 - p) = 44\times0.43\times0.57 \).

$$ 44\times0.43 = 18.92, \quad 18.92\times0.57 = 10.7844 $$

Now, take the square root: \( \sigma = \sqrt{10.7844} \approx 3.28 \) (rounded to two decimal places).

Answer:

The standard deviation is \( 3.28 \)