Question 186936
As an experiment, a self-confessed connoisseur of cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the unpopped kernels were counted. There were 86. 
(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
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sample proportion: 86/773 = 0.11 ;Sample standard deviation: sqrt(0.11*0.89/773)
= 0.011
E = 1.645*0.0113 = 0.0186
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90% CI: 0.11-0.0186 < p < 0.11+0.0186

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(b) Check the normality assumption.
pn >5 and qn >5
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(c) Try the Very Quick Rule.
I'm not sure what that is.
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Does it work well here? 
Why, or why not?
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(d) Why might this sample not be typical?
The sample might not be representative of the population.
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Cheers,
Stan H.