Question

In one city, 85% of eligible voters are registered to vote. If 425 randomly selected
eligible voters are surveyed, what is the probability that at least 82% of those selected are
registered to vote?

A food snack manufacturer samples 20 bags of pretzels off the assembly line and
weighed their contents. If the sample mean is 12.3 and the sample standard deviation is
0.24, find the 90% confidence interval of the true mean.

A random sample of 90 voters found that 43% were going to vote for a certain candidate.
Find the 99% confidence interval for the population proportion of voters who will vote for
that candidate.

Answer 1

1)

Here, μ = 0.85, σ = 0.0173 and x = 0.82. We need to compute P(X >= 0.82). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (0.82 - 0.85)/0.0173 = -1.73

Therefore,
P(X >= 0.82) = P(z <= (0.82 - 0.85)/0.0173)
= P(z >= -1.73)
= 1 - 0.0418 = 0.9582

2)

sample mean, xbar = 12.3
sample standard deviation, s = 0.24
sample size, n = 20
degrees of freedom, df = n - 1 = 19

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.729

ME = tc * s/sqrt(n)
ME = 1.729 * 0.24/sqrt(20)
ME = 0.0928

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (12.3 - 1.729 * 0.24/sqrt(20) , 12.3 + 1.729 * 0.24/sqrt(20))
CI = (12.21 , 12.39)

3)

sample proportion, = 0.43
sample size, n = 90
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.43 * (1 - 0.43)/90) = 0.0522

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

Margin of Error, ME = zc * SE
ME = 2.58 * 0.0522
ME = 0.135

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.43 - 2.58 * 0.0522 , 0.43 + 2.58 * 0.0522)
CI = (0.2953 , 0.5647)