Question 2: Central Limit Theorem Friends was a popular television show in the e

Question

Question 2: Central Limit Theorem Friends was a popular television show in the early 2000s and late 1990s. Nielsen a company that calculates television ratings surveyed 100 households and found that 36% of households watched the series finale of Friends. Let's represent watching the series finale as a random variable X so that X-1 means the household watched friends. a. Express the fraction of households in the sample that watched Friends as a sample statistic. b. What does the Central Limit Theorem tell us about the distribution of your statistic from part b? c. What is the mean and variance of your statistic from part a, in terms of p, the probability a single household watched the Friends finale. e. What is the probability that our sample statistic would be 36% or less if in fact the same percent of households watched the Friends finale as did the Seinfeld finale. Assume that 40% of households watched the Seinfeld finale. f. How many people would the survey need to survey in order for there only to be a 1% chance of getting 36% or less given that 40% of households watched the finale?

Explanation / Answer

a) Let n = number of household surveyed = 100

X = Number of household watched Friends

The sample statistic is sample propertion

p = sample propertion of household that watched friends = X/n = 36% = 0.36

b) By central limit theorem the sampling distribution of sample propertion is

Z = (p -E(p)) / S.D(p) ~ N(0,1)

c) Since X ~ Bin (n,p) then E(X) =np and var(X) =npq

p= X/n

Mean = E(p) = E(X/n) = p

and Variance = Var(p) = Var(X/n) = (1/n2) * Var(X) = pq/n

e) Required proability =P(p <=0.36 / p=0.40)

E(p) = 0.40 and Var(p) = 0.40 *0.60 /100 = 0.0024

S.D(p) = sqrt(var) =0.04899

P (p<=0.36) = P( (p-E(p))/S.D(p) < = (0.36 -0.40) /0.04899)

= P(Z < = -0.8165)

from normal probability table

P(Z <= -0.8165) = 0.2070

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