Question: The mass of a species of mouse commonly found in houses is normally distributed with a mean of $20.2$ grams with a standard deviation of $0.13$ grams. For parts (a) through (c), enter your responses as a decimal with 4 decimal places. a) What is the probability that a randomly chosen mouse has a mass of less than $20.05$ grams? b) What is the probability that a randomly chosen mouse has a mass of more than $20.31$ grams? c) What proportion of mice have a mass between $20.11$ and $20.29$ grams? d) $10\%$ of all mice have a mass of less than grams.

1. **Problem Statement:** We have a normally distributed variable $X$ representing mouse mass with mean $\mu = 20.2$ grams and standard deviation $\sigma = 0.13$ grams. 2. **Formula and Rules:** The standard normal variable $Z$ is given by: $$Z = \frac{X - \mu}{\sigma}$$ We use the standard normal distribution table or a calculator to find probabilities. --- ### Part (a): Probability $X < 20.05$ 3. Calculate the $Z$-score: $$Z = \frac{20.05 - 20.2}{0.13} = \frac{-0.15}{0.13} = -1.1538$$ 4. Find $P(Z < -1.1538)$ using standard normal tables or calculator: $$P(Z < -1.1538) = 0.1244$$ --- ### Part (b): Probability $X > 20.31$ 5. Calculate the $Z$-score: $$Z = \frac{20.31 - 20.2}{0.13} = \frac{0.11}{0.13} = 0.8462$$ 6. Find $P(Z > 0.8462)$: $$P(Z > 0.8462) = 1 - P(Z < 0.8462) = 1 - 0.8016 = 0.1984$$ --- ### Part (c): Proportion between $20.11$ and $20.29$ 7. Calculate $Z$-scores: $$Z_1 = \frac{20.11 - 20.2}{0.13} = \frac{-0.09}{0.13} = -0.6923$$ $$Z_2 = \frac{20.29 - 20.2}{0.13} = \frac{0.09}{0.13} = 0.6923$$ 8. Find probabilities: $$P(Z < 0.6923) = 0.7557$$ $$P(Z < -0.6923) = 0.2443$$ 9. Proportion between: $$P(-0.6923 < Z < 0.6923) = 0.7557 - 0.2443 = 0.5114$$ --- ### Part (d): Find $x$ such that $P(X < x) = 0.10$ 10. Find $Z$-score for $P(Z < z) = 0.10$: $$z = -1.2816$$ 11. Convert back to $X$: $$x = \mu + z \sigma = 20.2 + (-1.2816)(0.13) = 20.2 - 0.1666 = 20.0334$$ --- **Final answers:** - (a) $0.1244$ - (b) $0.1984$ - (c) $0.5114$ - (d) $20.0334$ grams