1. **Problem statement:** We have a normal distribution for Zaple smartphone battery lifetime $X$ with mean $\mu=8$ hours and standard deviation $\sigma=1.5$ hours. (a)(i) Find $P(X \neq 8)$. (a)(ii) Find $P(6 < X < 10)$. (b) Find the lifetime exceeded by 90% of Zaple batteries. (c) For Kaphone battery lifetime $Y \sim N(7, \sigma^2)$, find $\sigma$ given $P(Y < 5) = 0.25$. --- 2. **Recall properties of normal distribution:** - The probability that a continuous random variable equals a single point is zero, so $P(X = 8) = 0$. - Standardize using $Z = \frac{X - \mu}{\sigma}$. - Use standard normal tables or inverse CDF for probabilities and quantiles. --- 3. **(a)(i) Calculate $P(X \neq 8)$:** Since $X$ is continuous, $$P(X \neq 8) = 1 - P(X = 8) = 1 - 0 = 1.$$ --- 4. **(a)(ii) Calculate $P(6 < X < 10)$:** Standardize bounds: $$Z_1 = \frac{6 - 8}{1.5} = \frac{-2}{1.5} = -\frac{4}{3} = -1.333...$$ $$Z_2 = \frac{10 - 8}{1.5} = \frac{2}{1.5} = \frac{4}{3} = 1.333...$$ Using symmetry of normal distribution, $$P(6 < X < 10) = P(-1.333 < Z < 1.333) = 2 \times P(0 < Z < 1.333).$$ From standard normal tables, $P(Z < 1.333) \approx 0.9082$, so $$P(0 < Z < 1.333) = 0.9082 - 0.5 = 0.4082.$$ Therefore, $$P(6 < X < 10) = 2 \times 0.4082 = 0.8164.$$ --- 5. **(b) Find lifetime exceeded by 90% of batteries:** We want $x$ such that $$P(X > x) = 0.9 \implies P(X \leq x) = 0.1.$$ Find $z$ where $P(Z \leq z) = 0.1$. From tables, $z \approx -1.28155$. Convert back to $x$: $$x = \mu + z \sigma = 8 + (-1.28155)(1.5) = 8 - 1.9223 = 6.0777.$$ So, 90% of batteries last longer than approximately 6.08 hours. --- 6. **(c) Find $\sigma$ for Kaphone batteries:** Given $Y \sim N(7, \sigma^2)$ and $$P(Y < 5) = 0.25.$$ Standardize: $$P\left(Z < \frac{5 - 7}{\sigma}\right) = 0.25.$$ From standard normal tables, $P(Z < -0.6745) = 0.25$, so $$\frac{5 - 7}{\sigma} = -0.6745 \implies \frac{-2}{\sigma} = -0.6745.$$ Multiply both sides by $\sigma$ and divide by 0.6745: $$\cancel{\sigma} \times \frac{-2}{\cancel{\sigma}} = -0.6745 \implies -2 = -0.6745 \sigma$$ $$\Rightarrow \sigma = \frac{2}{0.6745} = 2.965.$$ Rounded to three significant figures, $\sigma = 2.97$. --- **Final answers:** (a)(i) $P(X \neq 8) = 1$ (a)(ii) $P(6 < X < 10) \approx 0.816$ (b) Lifetime exceeded by 90% is approximately 6.08 hours (c) $\sigma$ for Kaphone batteries is approximately 2.97 hours