1. **Problem statement:** We have weights of 10-year-old boys normally distributed with mean $\mu=44$ kg and standard deviation $\sigma=5$ kg. We want to find probabilities for weights between 43 and 45 kg for different sample sizes. 2. **Formula and rules:** For a normal distribution, the probability that a value $X$ lies between $a$ and $b$ is given by $$P(a < X < b) = P\left(\frac{a-\mu}{\sigma} < Z < \frac{b-\mu}{\sigma}\right)$$ where $Z$ is a standard normal variable with mean 0 and standard deviation 1. For sample means of size $n$, the sampling distribution has mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$. 3. **Part (a): Single boy weight between 43 and 45 kg** - Mean $\mu=44$, standard deviation $\sigma=5$ - Calculate $Z$-scores: $$Z_1 = \frac{43-44}{5} = \frac{-1}{5} = -0.2$$ $$Z_2 = \frac{45-44}{5} = \frac{1}{5} = 0.2$$ - Probability: $$P(43 < X < 45) = P(-0.2 < Z < 0.2) = \Phi(0.2) - \Phi(-0.2)$$ - Using symmetry $\Phi(-0.2) = 1 - \Phi(0.2)$, so $$P = 2\Phi(0.2) - 1$$ - From standard normal tables, $\Phi(0.2) \approx 0.5793$ - Thus, $$P = 2 \times 0.5793 - 1 = 0.1586$$ 4. **Part (b): Mean weight of 4 boys between 43 and 45 kg** - Sample size $n=4$ - Standard deviation of sample mean: $$\sigma_{\bar{X}} = \frac{5}{\sqrt{4}} = \frac{5}{2} = 2.5$$ - Calculate $Z$-scores: $$Z_1 = \frac{43-44}{2.5} = -0.4$$ $$Z_2 = \frac{45-44}{2.5} = 0.4$$ - Probability: $$P = \Phi(0.4) - \Phi(-0.4) = 2\Phi(0.4) - 1$$ - From tables, $\Phi(0.4) \approx 0.6554$ - Thus, $$P = 2 \times 0.6554 - 1 = 0.3108$$ 5. **Part (c): Mean weight of 25 boys between 43 and 45 kg** - Sample size $n=25$ - Standard deviation of sample mean: $$\sigma_{\bar{X}} = \frac{5}{\sqrt{25}} = \frac{5}{5} = 1$$ - Calculate $Z$-scores: $$Z_1 = \frac{43-44}{1} = -1$$ $$Z_2 = \frac{45-44}{1} = 1$$ - Probability: $$P = \Phi(1) - \Phi(-1) = 2\Phi(1) - 1$$ - From tables, $\Phi(1) \approx 0.8413$ - Thus, $$P = 2 \times 0.8413 - 1 = 0.6826$$ **Final answers:** - (a) $0.1586$ - (b) $0.3108$ - (c) $0.6826$