In a recent year, the scores for the reading portion of a test were normally distributed, with a mean of 21.2 and a standard deviation of 6.4. Complete parts (a) through (d) below (a) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 20. The probability of a student scoring less than 20 is 0.4251 (Round to four decimal places as needed) (b) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is between 14.0 and 28,4 The probability of a student sconing between 140 and 28.4 is 0.7394 (Round to four decimal places as needed) (c) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is more than 34.3 The probability of a student scoring more than 34.3 is \( \square \) (Round to four decimal places as needed)

To solve part (c) of the problem, we need to find the probability that a randomly selected high school student has a score greater than 34.3. We will use the properties of the normal distribution, specifically the z-score formula, to find this probability. ### Step 1: Calculate the z-score The z-score is calculated using the formula: \[ z = \frac{X - \mu}{\sigma} \] where: - \(X\) is the score we are interested in (34.3), - \(\mu\) is the mean (21.2), - \(\sigma\) is the standard deviation (6.4). ### Step 2: Substitute the values into the formula Let's calculate the z-score for \(X = 34.3\): \[ z = \frac{34.3 - 21.2}{6.4} \] Now, I will calculate the z-score. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(34.3-21.2\right)}{6.4}\) - step1: Remove the parentheses: \(\frac{34.3-21.2}{6.4}\) - step2: Subtract the numbers: \(\frac{13.1}{6.4}\) - step3: Convert the expressions: \(\frac{\frac{131}{10}}{\frac{32}{5}}\) - step4: Multiply by the reciprocal: \(\frac{131}{10}\times \frac{5}{32}\) - step5: Reduce the numbers: \(\frac{131}{2}\times \frac{1}{32}\) - step6: Multiply the fractions: \(\frac{131}{2\times 32}\) - step7: Multiply: \(\frac{131}{64}\) It appears that I cannot directly calculate the cumulative probability for the z-score. Instead, I will use a standard normal distribution table or a calculator to find \( P(Z < 2.0469) \). ### Step 4: Look up the cumulative probability Using a standard normal distribution table or calculator, we find: \[ P(Z < 2.0469) \approx 0.9798 \] ### Step 5: Calculate the probability of scoring more than 34.3 Now we can calculate \( P(Z > 2.0469) \): \[ P(Z > 2.0469) = 1 - P(Z < 2.0469) = 1 - 0.9798 = 0.0202 \] ### Final Answer The probability that a randomly selected high school student who took the reading portion of the test has a score that is more than 34.3 is approximately: \[ \boxed{0.0202} \] (Rounded to four decimal places as needed.)