Answer:
a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.
b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.
Step-by-step explanation:
For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
70% of adult women visit their doctors for a physical examination at least once in two years.
This means that
20 adult women
This means that
(a) Fewer than 14 of them have had a physical examination in the past two years.
This is:
In which
So
0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.
(b) At least 17 of them have had a physical examination in the past two years
From the values found in item (a).
0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.