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Wize High School Algebra II Textbook (Common Core) > Sampling and Estimating

Sampling Distribution for a Proportion

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Sampling Distribution of Proportions

By the Central LImit Theorem, if the sample size nn is large enough, then the sampling distribution for a proportion p^\hat{p} is approximately normal with mean pp and standard deviation σp^=pqn\large\sigma_{\hat{p}}=\large\sqrt{\frac{pq}{n}} , where,
  • pp is the population proportion (the parameter).
  • q=1−pq=1-p

We can then use the standardization formula for normal distribution to find the z-score, which will help us more easily calculate probabilities!
Z=p^−ppqn\boxed{\large Z=\frac{\hat{p}-p}{\large\sqrt{\frac{\large pq}{\large n}}}}

When a randomly selected sample size nn is drawn from a population, the sample proportion p^\hat{p} is denoted:

p^=Xn\large\hat{p}=\frac{X}{n}

0≤p^≤10\leq\hat{p}\leq1

where X is the number of individuals with a certain characteristic.

When is the Sample Size "Large Enough"?

The sample size is large enough if np ≥ 9   and   n(1−p) ≥ 9\boxed{np\ \geq\ 9~~~\text{and}~~~n(1-p)\ \geq\ 9}


Example

A report from 2008 revealed that 12% of undergraduate students avoid enrolling in weekend courses. Based on a recent survey of 70 undergraduate students, 11 of the say that they avoid enrolling in weekend courses. Is the sample size large enough?

np=(70)(0.12)=8.4<9 → NOnp=\left(70\right)\left(0.12\right)=8.4<9\ \rightarrow\ NO
nq=(70)(1−0.12)=(70)(0.88)=61.6>9 → YESnq=\left(70\right)\left(1-0.12\right)=\left(70\right)\left(0.88\right)=61.6>9\ \rightarrow\ YES

Sample is not large enough.
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Example: Sampling Distribution for a Proportion


Landlord Magazine revealed that 85% of renters pay their rents on time. Based on a random sample of 60 renters, what is the probability that more than 55 of them pay their rents on time?




p^=5560=0.9167\hat{p}=\dfrac{55}{60}=0.9167
z=p^−ppqnz=\dfrac{\hat{p}-p}{\sqrt{\dfrac{pq}{n}}}

z=0.9167−0.85(0.85)(0.15)60=1.45z=\dfrac{0.9167-0.85}{\sqrt{\dfrac{\left(0.85\right)\left(0.15\right)}{60}}}=1.45

Using the z-table:

P(z>1.45)=1−P(z<1.45)=1−0.9265=0.0735P\left(z>1.45\right)=1-P(z<1.45)=1-0.9265=0.0735


Miles claims that 4% of sales get refunded. You randomly select 200 transactions. What is the probability that more than 12 transactions in his sample got refunded?