1. Problem: A man tosses a fair coin 196 times. Find the probability that he obtains between 105 and 112 heads. Step 1: Define the problem using the binomial distribution with parameters $n=196$ and $p=0.5$. Step 2: Use the normal approximation to the binomial since $n$ is large. The mean is $\mu = np = 196 \times 0.5 = 98$ and the standard deviation is $\sigma = \sqrt{np(1-p)} = \sqrt{196 \times 0.5 \times 0.5} = 7$. Step 3: Apply continuity correction: find $P(104.5 < X < 112.5)$. Step 4: Convert to standard normal variable $Z = \frac{X - \mu}{\sigma}$. Calculate $Z_1 = \frac{104.5 - 98}{7} = 0.93$ and $Z_2 = \frac{112.5 - 98}{7} = 2.07$. Step 5: Use standard normal tables or calculator to find $P(0.93 < Z < 2.07) = \Phi(2.07) - \Phi(0.93) \approx 0.9808 - 0.8238 = 0.157$. 2. Problem: A man tosses a coin 196 times and obtains 105 heads. At 5% significance, test if the coin is fair. Step 1: Null hypothesis $H_0: p=0.5$, alternative $H_a: p \neq 0.5$. Step 2: Use normal approximation with $\mu=98$, $\sigma=7$. Step 3: Calculate test statistic $Z = \frac{105 - 98}{7} = 1$. Step 4: Critical values at 5% two-tailed test are $\pm 1.96$. Step 5: Since $|1| < 1.96$, fail to reject $H_0$. The coin is fair at 5% level. 3. Problem: In a test, 20/50 boys pass and 30/50 girls pass. At 5% significance, test if performance differs. Step 1: Define $p_1=20/50=0.4$, $p_2=30/50=0.6$, $n_1=n_2=50$. Step 2: Null hypothesis $H_0: p_1 = p_2$, alternative $H_a: p_1 \neq p_2$. Step 3: Pooled proportion $p = \frac{20+30}{50+50} = 0.5$. Step 4: Standard error $SE = \sqrt{p(1-p)(\frac{1}{n_1} + \frac{1}{n_2})} = \sqrt{0.5 \times 0.5 \times (\frac{1}{50} + \frac{1}{50})} = 0.1$. Step 5: Test statistic $Z = \frac{0.4 - 0.6}{0.1} = -2$. Step 6: Critical values $\pm 1.96$. Since $|-2| > 1.96$, reject $H_0$. There is a significant difference. 4. Problem: IQ test variances: boys 10 (n=16), girls 15 (n=25). Test if variances differ at 5%. Step 1: Null hypothesis $H_0: \sigma_1^2 = \sigma_2^2$, alternative $H_a: \sigma_1^2 \neq \sigma_2^2$. Step 2: Calculate $F = \frac{s_2^2}{s_1^2} = \frac{15}{10} = 1.5$ (larger variance over smaller). Step 3: Degrees of freedom $df_1=24$, $df_2=15$. Step 4: Critical values from F-table at 5% two-tailed test approximately $F_{0.025,24,15} \approx 2.54$ and $F_{0.975,24,15} \approx 0.39$. Step 5: Since $0.39 < 1.5 < 2.54$, fail to reject $H_0$. Variances are not significantly different. 5. Problem: Maths class means: boys 65 (sd=4, n=11), girls 60 (sd=5, n=11). Test if boys perform better at 5%. Step 1: Null hypothesis $H_0: \mu_1 = \mu_2$, alternative $H_a: \mu_1 > \mu_2$. Step 2: Calculate pooled standard deviation: $$s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}} = \sqrt{\frac{10 \times 16 + 10 \times 25}{20}} = \sqrt{20.5} = 4.527$$ Step 3: Calculate test statistic: $$t = \frac{65 - 60}{s_p \sqrt{\frac{1}{11} + \frac{1}{11}}} = \frac{5}{4.527 \times 0.426} = \frac{5}{1.93} = 2.59$$ Step 4: Degrees of freedom $df=20$, critical t-value at 5% one-tailed is about 1.725. Step 5: Since $2.59 > 1.725$, reject $H_0$. Boys perform better. 6. Problem: Weight of 5000 bags normally distributed with mean 100. Sample of 25 bags has mean 98, sd 5. Test if bags are underweight at 5%. Step 1: Null hypothesis $H_0: \mu = 100$, alternative $H_a: \mu < 100$. Step 2: Calculate test statistic: $$t = \frac{98 - 100}{5 / \sqrt{25}} = \frac{-2}{1} = -2$$ Step 3: Degrees of freedom $df=24$, critical t-value at 5% one-tailed is about -1.711. Step 4: Since $-2 < -1.711$, reject $H_0$. Bags are underweight. 7. Problem: Marks normal with mean 66, sd 6. Find probabilities: 7.1 Probability mark less than 60: $$Z = \frac{60 - 66}{6} = -1$$ $$P(X < 60) = \Phi(-1) = 0.1587$$ 7.2 Probability mark greater than 60: $$P(X > 60) = 1 - 0.1587 = 0.8413$$ 7.3 Probability mark between 57 and 75: $$Z_1 = \frac{57 - 66}{6} = -1.5, \quad Z_2 = \frac{75 - 66}{6} = 1.5$$ $$P(57 < X < 75) = \Phi(1.5) - \Phi(-1.5) = 0.9332 - 0.0668 = 0.8664$$ 7.4 Probability mark between 57 and 63: $$Z_1 = \frac{57 - 66}{6} = -1.5, \quad Z_2 = \frac{63 - 66}{6} = -0.5$$ $$P(57 < X < 63) = \Phi(-0.5) - \Phi(-1.5) = 0.3085 - 0.0668 = 0.2417$$