We are interested in conducting a study in order to determine what percentage of voters of a state would vote for the incumbent governor. what is the minimum size sample needed to estimate the population proportion with a margin of error of 0.05 or less at 95% confidence?

200

100

58

196

385

Step-by-step explanation:

Lets say that p is the probability for the governor to be voted by a person.

Lets call X the random variable with value 1 if a person votes for the governor and 0 if not. X is a random variable with Bernoulli distribution of parameter p. X has mean p and variance p(1-p). Lets denote Y the mean of a sample of lenght n. Y has mean μ=p and variance σ² = p*(1-p)/n.

As a consecuence of the Central Limit Theorem, since Y is the mean of identically distributed independent random variables, we have that Y is approximately Normal, with parameters μ = p and σ = √(p(1-p)/n).

As a consecuence of the Law of Big Numbers, we have that  

The values of Ф, the cummulative function of the normal density funtion, can be obtained from the attached file. We want a 95% confidence interval, so we want Z that satisfies

P(-Z < W < Z) = 0.95

Thus, P(W < Z) = 0.975

If we look the table, we will find that Z = 1.96, Hence

Now we move the Y and the - sign, that changes the inequalities, so we obtain

Therefore, the confidence interval is , which has lenght . In the worst case, Y(1-Y) takes the value 0.25 (when the sample proportion is 1/2). Thus, the lenght of the interval is bounded above by

In order for the lenght to be at most 0.1 (so, the margin of error is 0.05) we need n such that

Hence,

We take n = 385. None of the 4 given values would work according to the information we have. A sample size of lenght 200 would give us a confidence interval of lenght almost 0.2

i believe the answer is 68cm

step-by-step explanation:

i just took 34 x 2 because the space between v and y is the same as y to k.