1. **State the problem:** We are given the probability statement $$P(0.2 - a < \hat{P}_1 - \hat{P}_2 < 0.2 + a) = 0.1551$$ where $$P_1 = 0.25$$, $$P_2 = 0.2$$, $$n_1 = 5$$, and $$n_2 = 10$$. We need to find the value of $$a$$. 2. **Recall the formula for the difference of two sample proportions:** $$\hat{P}_1 - \hat{P}_2 \sim N\left(P_1 - P_2, \sqrt{\frac{P_1(1-P_1)}{n_1} + \frac{P_2(1-P_2)}{n_2}}\right)$$ 3. **Calculate the mean difference:** $$P_1 - P_2 = 0.25 - 0.2 = 0.05$$ 4. **Calculate the standard deviation (standard error) of the difference:** $$\sigma = \sqrt{\frac{0.25 \times 0.75}{5} + \frac{0.2 \times 0.8}{10}} = \sqrt{\frac{0.1875}{5} + \frac{0.16}{10}} = \sqrt{0.0375 + 0.016} = \sqrt{0.0535} \approx 0.2313$$ 5. **Rewrite the probability in terms of the standard normal variable:** Let $$Z = \frac{(\hat{P}_1 - \hat{P}_2) - (P_1 - P_2)}{\sigma}$$ The probability becomes: $$P\left(\frac{0.2 - a - 0.05}{0.2313} < Z < \frac{0.2 + a - 0.05}{0.2313}\right) = 0.1551$$ Simplify the bounds: $$P\left(\frac{0.15 - a}{0.2313} < Z < \frac{0.15 + a}{0.2313}\right) = 0.1551$$ 6. **Use symmetry and standard normal distribution properties:** The interval is symmetric around $$\frac{0.15}{0.2313} \approx 0.648$$, so the probability between $$0.648 - \frac{a}{0.2313}$$ and $$0.648 + \frac{a}{0.2313}$$ is 0.1551. 7. **Find the z-values corresponding to the probability 0.1551:** Since the probability is small and centered at 0.648, we find the half-width $$h = \frac{a}{0.2313}$$ such that: $$P(0.648 - h < Z < 0.648 + h) = 0.1551$$ This means: $$\Phi(0.648 + h) - \Phi(0.648 - h) = 0.1551$$ 8. **Calculate the cumulative probabilities:** Let $$p_1 = \Phi(0.648 - h)$$ and $$p_2 = \Phi(0.648 + h)$$, so $$p_2 - p_1 = 0.1551$$. 9. **Use numerical or inverse normal calculations:** Using symmetry and the standard normal table or calculator, find $$h$$ such that the above holds. 10. **Approximate solution:** By trial or using a calculator, $$h \approx 0.12$$. 11. **Solve for $$a$$:** $$a = h \times 0.2313 \approx 0.12 \times 0.2313 = 0.0278$$ **Final answer:** $$a \approx 0.028$$