An article stated, \"Surveys tell us that more than half of America\'s college g

Question

An article stated, "Surveys tell us that more than half of America's college graduates are avid readers of mystery novels." Let p denote the actual proportion of college graduates who are avid readers of mystery novels. Consider a sample proportion p that is based on a random sample of 235 college graduates.

(a) If p = 0.5, what are the mean value and standard deviation of p? (Round your answers to four decimal places.)

If p = 0.6, what are the mean value and standard deviation of p? (Round your answers to four decimal places.)

Does p have approximately a normal distribution in both cases? Explain.

Yes, because in both cases np > 10 and n(1 p) > 10.No, because in both cases np < 10 or n(1 p) < 10.    No, because when p = 0.5, np < 10.No, because when p = 0.6, np < 10.

(b) Calculate P(p 0.6) for p = 0.5. (Round your answer to four decimal places.)


Calculate P(p 0.6) for p = 0.6.


(c) Without doing any calculations, how do you think the probabilities in Part (b) would change if n were 410 rather than 235?

When p = 0.5, the P(p 0.6) would remain the same if the sample size was 410 rather than 235. When p = 0.6, the P(p 0.6) would remain the same if the sample size was 410 rather than 235.When p = 0.5, the P(p 0.6) would decrease if the sample size was 410 rather than 235. When p = 0.6, the P(p 0.6) would remain the same if the sample size was 410 rather than 235.    When p = 0.5, the P(p 0.6) would remain the same if the sample size was 410 rather than 235. When p = 0.6, the P(p 0.6) would decrease if the sample size was 410 rather than 235.When p = 0.5, the P(p 0.6) would decrease if the sample size was 410 rather than 235. When p = 0.6, the P(p 0.6) would decrease if the sample size was 410 rather than 235.

Explanation / Answer

a) for p=0.5

mean =0.5

std deviation =(p(1-p)/n)1/2 =0.0326

for p=0.6

mean =0.6

std deviation==(p(1-p)/n)1/2 =0.0320

Yes, because in both cases np > 10 and n(1 p) >10

b)

for p=0.5

P(p 0.6) =P(Z>(0.6-0.5)/0.0326)=P(Z>3.0659)=1-P(Z<3.0659)=1-0.9989=0.0011

for p=0.6

P(p 0.6) =P(Z>(0.6-0.5)/0.0320)=P(Z>0)=0.5

c)

.When p = 0.5, the P(p 0.6) would decrease if the sample size was 410 rather than 235. When p = 0.6, the P(p 0.6) would remain the same if the sample size was 410 rather than 235

please revert for any clarification required,

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