Estate Planning as a Career

What is the mean number of smokers in the sample?what is standard deviation?

March 31st, 2011 Filed under: sample wills — Estate Planning Author

manu asks the question: In a large orgainisation 20% of the employees are smokers .A random sample of 400 employees is selected for the purpose of the studey?
1.What is the mean number of smolkers in the sample?What is the standard deviation?
2.Use the normal approximation to calculate the probability that the sample will contain at least 70 smokers

Answer:This is a binomial random variable with parameters n=400 and p=.20

μ = np = 400 × 1/5 = 80
σ² = np(1-p) = 400 × 1/5 × 4/5 = 64
σ = 8

So for the first question, the answer is:
μ=80 and σ = 8

Using the Central Limit Theorem, it can be approximated as normally distributed with the same mean (μ=80) and variance(σ²=64)..

When n is large, the Binomial distribution X and the Normal distribution Y are related by the following equation:

P[ j ≤ X ≤ k ] = P[ j-1/2 < Y < k+1/2 ]

(so set j=70 and k=400)

P[69.5 < Y < 400.5] = .905324 (using normal approximation)

P[70 ≤ X ≤ 400] = .906993 (exact using binomial distribution)

As seen here, the central limit theorem is a very good approximation for this size sample..

I used these formulas in MS excel to calculate the results for the normal and binomail distributions respectively

=normdist(400.5,80,8,true) – normdist(69.5,80,8,true)
=1 – binomdist(69,400,.2,true)(Higher Education (University +))