During the 2008 presidential election, newspapers reported that Obama received 40% of the white male vote. If this is true, what is the probability that in a random sample of 1,000 white male voters between 37% and 43% voted for Obama?  Do not write as a percent e.g. 95.57% should be entered as 0.9557. 

Round the standard error to 3 decimal places. 

This problem involves using a normal approximation to the binomial distribution because we are dealing with a large sample size of 1,000 voters.

### Step 1: Define the variables
- The sample size \( n = 1,000 \)
- The proportion of white male voters who voted for Obama \( p = 0.40 \)
- We are looking for the probability that between 37% and 43% of the voters (i.e., between 0.37 and 0.43) voted for Obama.

### Step 2: Calculate the standard error (SE)
The standard error of the proportion is calculated using the formula:

\[
SE = \sqrt{\frac{p(1-p)}{n}}
\]

Substitute \( p = 0.40 \), \( 1 - p = 0.60 \), and \( n = 1,000 \):

\[
SE = \sqrt{\frac{0.40 \times 0.60}{1,000}} = \sqrt{\frac{0.24}{1,000}} = \sqrt{0.00024} \approx 0.0155
\]

Thus, the standard error is \( SE \approx 0.016 \) (rounded to 3 decimal places).

### Step 3: Convert the percentages to z-scores
Now, we convert the given percentages (0.37 and 0.43) to z-scores using the formula:

\[
z = \frac{\hat{p} - p}{SE}
\]

- For \( \hat{p} = 0.37 \):

\[
z_1 = \frac{0.37 - 0.40}{0.016} = \frac{-0.03}{0.016} \approx -1.875
\]

- For \( \hat{p} = 0.43 \):

\[
z_2 = \frac{0.43 - 0.40}{0.016} = \frac{0.03}{0.016} \approx 1.875
\]

### Step 4: Find the probabilities
Using the z-table (or a standard normal distribution calculator), we find the probabilities corresponding to these z-scores:

- For \( z_1 = -1.875 \), the cumulative probability is approximately 0.0304.
- For \( z_2 = 1.875 \), the cumulative probability is approximately 0.9696.

### Step 5: Calculate the probability between 0.37 and 0.43
The probability that the proportion of voters is between 0.37 and 0.43 is:

\[
P(0.37 \leq \hat{p} \leq 0.43) = P(z_2) - P(z_1) = 0.9696 - 0.0304 = 0.9392
\]

Thus, the probability is approximately \( 0.9392 \).

### Final Answer:
The probability that between 37% and 43% of white male voters voted for Obama in a random sample of 1,000 voters is **0.9392**.

Would you like further details or clarification?

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Here are 5 related questions to explore:

1. How does increasing the sample size affect the standard error in this case?
2. What if the percentage of white male voters for Obama was 45% instead of 40%?
3. How would the probability change if we were interested in a narrower range, say between 38% and 42%?
4. What are the limitations of using a normal approximation for a binomial distribution?
5. How would you calculate the probability if the sample size was smaller, say 100 instead of 1,000?

**Tip:** When dealing with large sample sizes, the binomial distribution can often be approximated using the normal distribution, simplifying the calculations.

During the 2008 presidential election, newspapers reported that Obama received 40% of the white male vote. If this is true, what is the probability that in a random sample of 1,000 white male voters between 37% and 43% voted for Obama? Do not write as a percent e.g. 95.57% should be entered as 0.9557. Round the standard error to 3 decimal places.

Learn how to calculate the probability that between 37% and 43% of white male voters supported Obama in a sample of 1,000 during the 2008 election. This problem involves using normal approximation to the binomial distribution and calculating the standard error, z-scores, and probability. Suitable for students in Grades 10-12.

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