Question 506904
Biting an unpopped kernel of popcorn hurts! 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.
p-hat = 86/773 = 0.111..
E = 1.645*sqrt[(0.111)(0.889)/773]=0.0186
90% CI: 0.111-0.0186 < p < 0.111+0.0186
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(b) Check the normality assumption
I'll let you do that; it differs with every text.

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(c) Try the Very Quick Rule.
Does it work well here? Why, or why not?
I'm not sure what that is; it depends on your text.
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(d) Why might this sample not be typical?
The 773 is probably not a simple-random-sample
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Cheers,
Stan H.