Question

According to the Australian Bureau of Statistics, 70% of all Australian men aged 18 and over are overweight or obese. A student takes a random sample of 20 men aged 18 years and over. A particular variable of interest is the ‘number of men aged 18 and over who are overweight or obese’. Based on the above information answer the following questions:

(a) What is an appropriate model to represent the variable of interest? Write down the parameters of the model and their values, if any.

(b) Discuss how the conditions of the appropriate model are satisfied in the context of the current study.

(c) Find the mean and standard deviation of the number of men aged 18 years and over who are overweight or obese using the parameters of the model.

(d) Find the probability that at least 15 of the men aged 18 years and over are overweight or obese.

(e) Determine the probability that, in a random sample of 100 men aged 18 years or more, 75 or more men are overweight or obese. State and check any assumptions, conditions or rules of thumb that should be considered before performing the calculations to determine this probability.

Answer 1

A) The variable of interest represents a binomial model.

n = 20

p = 0.7

b) The experiment has n = 20 fixed number of trials.

Each trial is independent of the others.

There are only two outcomes. Probability of success = 0.7 and Probably of failure = 0.3

The probability of each outcome remains constant from trial to trial.

c) mean() = np = 20 * 0.7 = 14

Standard deviation() = sqrt(np(1 - p))

= sqrt(20 * 0.7 * 0.3)

= 2.0494

d) P(X > 15)

= P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

= 20C15 * (0.7)^15 * (0.3)^5 + 20C16 * (0.7)^16 * (0.3)^4 + 20C17 * (0.7)^17 * (0.3)^3 + 20C18 * (0.7)^18 * (0.3)^2 +20C19 * (0.7)^19 * (0.3)^1 +  20C20 * (0.7)^20 * (0.3)^0 = 0.4164

e) np = 100 * 0.7 = 70

n(1 - p) = 100 * (1 - 0.7) = 30

Since np > 5 and nq > 5, so we can use normal approximation to the binomial distribution.

= np = 100 * 0.7 = 70

= sqrt(np(1 - p))

= sqrt(100 * 0.7 * 0.3)

= 4.5826

P(X > 75)

= P((X - )/ > (74.5 - )/)

= P(Z > (74.5 - 70)/4.5826)

= P(Z > 0.982)

= 1 - P(Z < 0.982)

= 1 - 0.8370

= 0.1630