(Sampling Theory) This problem considers three different ways of answering a que

Question

(Sampling Theory) This problem considers three different ways of answering a question about samples from an infinite population. Suppose you flip a fair coin 120 times. What is the probability that 5/8's or more of the flips will be heads? (a) First solve this problem precisely using the binomial, explaining your procedure. (b) Next, solve the problem by using the normal approximation to the binomial (using the continuity correction), explaining your procedure. (c) Finally, solve it as a problem in sampling: consider the population to be normally distributed, and note that this is a question about proportions (so use the mean and standard deviation derived from the Bernoulli) and the sample size is 120. The sample distribution of means now gives us an estimate of the proportion of the population which is heads (which we know to be 0.5), the expected standard deviation, and we want to know the probability that the proportion (i.e., the mean) of the sample is 5/8's or more. Show all work.

Explanation / Answer

here n=120, 5/8 times more heard so number of heads=(5/8)*120=75

here P(head)=p=0.5

(a) P(X>=75)=1-P(X<74)=1-0.996=0.004

P(X<74)=0.996 ( using ms-excel=binomdist(74,120,0.5,1)

(b) here we use standard normal variate z=(x-mean)/sd

mean=n*p=120*0.5=60,

sd=sqrt(np(1-p))=sqrt(120*0.5*(1-0.5))=5.47

fpr x=75, z=(75-60)/5.47=2.74

P(X>=75)=P(Z>2.74)=1-P(Z<2.74)=1-0.997=0.03

(c) here we use standard normal varaite z=(p-P)/(sqrt(P(1-P)/n)=(5/8-0.5)/sqrt(0.5*(1-0.5)/120)=2.74

P(P>5/8)=P(Z>2.74)=1-P(Z<2.74)=1-0.997=0.003

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