Digression: Before introducing , we have used and s as statistics, and μ and σ as parameters. Often
we represent statistics with English letters and parameters with Greek letters. However, we depart from
that convention here by using as a statistic and p as a parameter. There are texts that are true to the
English/Greek convention by using p for the sample proportion and II for the population proportion.
We learned in the first section of this chapter that, if X is a binomial random variable, the mean and
standard deviation of the sampling distribution of X are given by
We know that if we divide each term in a dataset by the same value n , then the mean and standard
deviation of the transformed dataset will be the mean and standard deviation of the original dataset
divided by n . Doing the algebra, we find that the mean and standard deviation of the sampling
distribution of are given by:
Like the binomial, the sampling distribution of will be approximately normally distributed if n and p
are large enough. The test is exactly the same as it was for the binomial: If X has B (n, p ), and = X /n ,
then has approximately
provided that np ≥ 10 and n (1 – p ) ≥ 10 (or np ≥ 5 and n (1 – p) ≥ 5).
example: Harold fails to study for his statistics final. The final has 100 multiple- choice
questions, each with 5 choices. Harold has no choice but to guess randomly at all 100
questions. What is the probability that Harold will get at least 30% on the test?
solution: Since 100(0.2) and 100(0.8) are both greater than 10, we can use the normal
approximation to the sampling distribution of . Since
Therefore,
. The TI-83/84 solution is given by
normalcdf(0.3,100,0.2,0.040)=0.0062.
Harold should have studied.
Rapid Review
A coin is known to be unbalanced in such a way that heads only comes up 0.4 of the time.
(a) What is the probability the first head appears on the 4th toss?