1. **Problem statement:** We are given a mean systolic blood pressure $\mu = 128$ and a standard deviation $\sigma = 9$. We want to find the probability that a randomly selected person has a systolic blood pressure between 125 and 135. 2. **Formula and concept:** We use the normal distribution properties. 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. 3. **Calculate the Z-scores:** $$Z_1 = \frac{125 - 128}{9} = \frac{-3}{9} = -0.3333$$ $$Z_2 = \frac{135 - 128}{9} = \frac{7}{9} = 0.7778$$ 4. **Find probabilities from Z-table or standard normal distribution:** - $P(Z < -0.3333) \approx 0.3694$ - $P(Z < 0.7778) \approx 0.7818$ 5. **Calculate the probability between 125 and 135:** $$P(125 < X < 135) = P(Z < 0.7778) - P(Z < -0.3333) = 0.7818 - 0.3694 = 0.4124$$ 6. **Compare with given options:** The closest option to 0.4124 is option b. 0.4184. **Final answer:** The probability that a randomly selected person has systolic blood pressure between 125 and 135 is approximately **0.4184** (option b).