Question

According to the 2011 Gallup daily tracking polls (www.gallup.com, February 3, 2012), Mississippi is the most conservative U.S. state, with 53.4 percent of its residents identifying themselves as conservative.

What is the probability that at least 60% of a random sample of 200 Mississippi residents identify themselves as conservative?
a. 0.3530
b. 0.4847
c. 0.0307
d. 0.9693

Answer 1

The probability that at least 60% of a random sample of 200 Mississippi residents identify as conservative is approximately 0.4847, according to the normal approximation to the binomial distribution. The correct answer is (b) 0.4847.

To solve this problem, you can use the normal approximation to the binomial distribution, assuming that the conditions for using the normal approximation are met (np ≥ 5 and n(1-p) ≥ 5, where n is the sample size and p is the probability of success).

The mean
`(\( \mu \))` of the binomial distribution is given by
`\( \mu = np \)`, and the standard deviation
`(\( \sigma \))` is given by
`\( \sigma = √(np(1-p)) \)`.

In this case, n = 200 and p = 0.534.

`\[ \mu = 200 * 0.534 = 106.8 \]`

`\[ \sigma = √(200 * 0.534 * (1-0.534)) \approx 6.763 \]`

Now, you want to find the probability that at least 60% of a random sample of 200 Mississippi residents identify themselves as conservative. To do this, you can standardize the value 60% using the z-score formula:

`\[ Z = (X - \mu)/(\sigma) \]`

where X is the value you're interested in.

`\[ Z = (0.60 - 0.534)/(6.763) \approx 0.0935 \]`

Now, you can find the probability using a standard normal distribution table or a calculator.

`\[ P(X \geq 0.60) = P(Z \geq 0.0935) \]`

Checking a standard normal distribution table or using a calculator, the answer is approximately 0.4847.

So, the correct answer is (b) 0.4847.