Question

In Washington state, 68.2% of adults visit the dentist for their recommended twice-yearly cleanings. Let X be the random variable that counts the number of adults in a group of 5 people who have visited the dentist for their recommended twice-yearly cleanings.

(a) In the group of 5 people, find P(X=3), or the probability that exactly 3 in the group of 5 have visited the dentist for their twice-yearly cleaning. (round to 4 decimal places)

(b) In the group of 5 people, find the probability that at least 3 in the group of 5 have visited the dentist for their twice-yearly cleaning. (round to 4 decimal places)

(c) In the group of 5 people, find the probability that no more than 3 in the group of 5 have visited the dentist for their twice-yearly cleaning. (round to 4 decimal places)

Answer 1

Solution:

Given: In Washington state, 68.2% of adults visit the dentist for their recommended twice-yearly cleanings.

thus p = probability of adults visit the dentist for their recommended twice-yearly cleanings = 0.682

X be the random variable that counts the number of adults in a group of 5 people who have visited the dentist for their recommended twice-yearly cleanings.

Thus n = Number of adults selected = 5

Since X has two possible outcomes with constant probability of success and independent trials.

Thus X follows a binomial distribution with parameters n= 5 and p = 0.682

Part a) Find P(X = 3) = ......?

Binomial probability formula :

where

Thus

Part b) find the probability that at least 3 in the group of 5 have visited the dentist for their twice-yearly cleaning.

That is:

We know P(X =3) = 0.3208

Now find P(X= 4) and P(X = 5)

Thus

and

Thus

Part c) find the probability that no more than 3 in the group of 5 have visited the dentist for their twice-yearly cleaning.

P( No more than 3) =