A study reports that 36​% of companies in Country A have three or more female board directors. Suppose you select a random sample of 100 respondents. Complete parts​ (a) through​ (c) below.
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Part 1
a. What is the probability that the sample will have between 31​% and 43​% of companies in Country A that have three or more female board​ directors?
The probability is 
  
0.7787. ​(Round to four decimal places as​ needed.)
Part 2
b. The probability is 90​% that the sample percentage of Country A companies having three or more female board directors will be contained within what symmetrical limits of the population​ percentage?
The probability is 90​% that the sample percentage will be contained above 
  
enter your response here​% and below 
  
enter your response here​%

Let's solve this step-by-step.

### Part a. 

The solution is already provided as the probability of having between 31% and 43% of companies with three or more female board directors is \( 0.7787 \).

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### Part b. 

Now we need to calculate the **symmetrical limits** around the population percentage (36%) such that 90% of the sample proportions lie within those limits. This is essentially calculating a confidence interval for the sample proportion.

Here are the steps to calculate the 90% confidence interval:

1. **Identify the parameters:**
   - Population proportion \( p = 0.36 \)
   - Sample size \( n = 100 \)
   - Confidence level = 90%, so the significance level \( \alpha = 0.10 \). For a 90% confidence interval, the Z-value corresponding to \( \alpha/2 = 0.05 \) is \( Z = 1.645 \).

2. **Calculate the standard error (SE) for the sample proportion:**
   \[
   SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.36(1-0.36)}{100}} = \sqrt{\frac{0.36 \times 0.64}{100}} = \sqrt{0.2304 / 100} = 0.048
   \]

3. **Calculate the margin of error (ME):**
   \[
   ME = Z \times SE = 1.645 \times 0.048 = 0.07896 \approx 0.079
   \]

4. **Determine the confidence interval:**
   \[
   \text{Lower limit} = p - ME = 0.36 - 0.079 = 0.281
   \]
   \[
   \text{Upper limit} = p + ME = 0.36 + 0.079 = 0.439
   \]

5. **Convert the confidence interval to percentages:**
   - Lower limit: \( 0.281 \times 100 = 28.1\% \)
   - Upper limit: \( 0.439 \times 100 = 43.9\% \)

Thus, the probability is 90% that the sample percentage of Country A companies having three or more female board directors will be contained between **28.1% and 43.9%**.

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### Summary:
- **a.** The probability that the sample will have between 31% and 43% of companies is \( 0.7787 \).
- **b.** The probability is 90% that the sample percentage will be contained between **28.1% and 43.9%**.

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Would you like to go into more detail on any part? Here are some related questions:

1. What is the Z-value for a 95% confidence interval?
2. How do we interpret a confidence interval in terms of population proportions?
3. How does increasing the sample size affect the margin of error?
4. What assumptions are needed for this type of proportion analysis to be valid?
5. How would the calculations change if we wanted a 95% confidence interval?

**Tip:** The wider the confidence interval, the more uncertain we are about the true population proportion. Adjusting the confidence level or sample size can change this width.

A study reports that 36% of companies in Country A have three or more female board directors. Suppose you select a random sample of 100 respondents. Complete parts (a) through (c) below. a. What is the probability that the sample will have between 31% and 43% of companies in Country A that have three or more female board directors? The probability is 0.7787. (Round to four decimal places as needed.) b. The probability is 90% that the sample percentage of Country A companies having three or more female board directors will be contained within what symmetrical limits of the population percentage?

Learn how to calculate the probability and confidence intervals for the percentage of companies with three or more female board directors in a random sample of 100 respondents from Country A. This problem involves concepts such as probability, confidence intervals, and the central limit theorem. Suitable for Grades 11-12 and college-level statistics.

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