Answer to Question #128203 in Statistics and Probability for Abby

Question #128203
Suppose a simple random sample of size n=50 is taken from a population of N=15,000 with a population proportion of p=0.4 having a certain characteristic. What is the probability that a simple random sample will have x > 21 having that characteristic?

(Round to four decimal places)
1
Expert's answer
2020-08-03T18:43:20-0400

Check that the sample size is large enough:


np=500.4=2010n\cdot p=50\cdot0.4=20\geq10

np(1p)=500.4(10.4)=1210np(1-p)=50\cdot0.4(1-0.4)=12\geq10

Sampling distribution for p^\hat{p} is approximately normal:


μp^=p=0.4\mu_{\hat{p}}=p=0.4

σp^=p(1p)n=0.4(10.4)500.069282\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.4(1-0.4)}{50}}\approx0.069282

2150=0.42{21\over 50}=0.42

P(p^>0.42)=1P(p^0.42)=P(\hat{p}>0.42)=1-P(\hat{p}\leq0.42)=

=1P(Z0.420.40.069282)1P(Z0.288675)=1-P(Z\leq\dfrac{0.42-0.4}{0.069282})\approx1-P(Z\leq0.288675)\approx

0.386415\approx0.386415

The probability that a simple random sample will have x > 21 having that characteristic is approximately 0.3864.0.3864.



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