normal distribution to approximate a binomial distribution.

Question

normal distribution to approximate a binomial distribution.


2,881 adults were selected for a survey. if the probability, p, of a response to the survey is 71.85% what would be the probability that 2,086 or less adults would respond to the survey?

Answer 1

We need the mean and standard deviation number of responders.

 

Recall that for the Binomial distribution, they are μ=np and σ=√[np(1-p)]

 

Here we have n=2881 and p=0.7185 so μ=(2881)(0.7185)=2070 and σ=√[(2881)(0.7185)(1-0.7185)]=24.14

 

The z-score is z=(x-μ)/σ=(2086-2070)/24.14=0.6628

 

Using the standard normal distribution table or a statistics software gives a probability of 0.7463 or about 75% that there will be at most 2086 responders.