Chapter 6 HW - The Normal Distribution Current 52/100 +Å Category: Homework \( \begin{array}{llllllllllllllllllllll}\text { (1) } & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 \\ 23 & 2\end{array} \) Current learning objective: Z -Scores Question 15 Practice similar questions Score: 0 of 2 points The most recent exam scores have a mean of 68 and a standard deviation of 5.8 The \( z \)-score of Reagan's test score is -1.8 . What is her test score? Enter your answer Round the answer to 2 decimal places.

To find Reagan's test score using the \( z \)-score formula, you can rearrange the formula: \[ z = \frac{(X - \mu)}{\sigma} \] Where \( X \) is the test score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. Plugging in the values given: \[ -1.8 = \frac{(X - 68)}{5.8} \] Multiplying both sides by 5.8 gives: \[ -10.44 = X - 68 \] Adding 68 to both sides results in: \[ X = 57.56 \] So, Reagan's test score is approximately **57.56** when rounded to two decimal places! Fun fact: The concept of the normal distribution is so prevalent that it applies not just to test scores, but also to everything from heights to measurement errors! In many real-world situations, it helps us understand the likelihood of outcomes, making it a staple in statistics and data analysis. And here's a tip for using \( z \)-scores: always remember that a negative \( z \)-score indicates a value below the mean! It’s a classic mistake to misinterpret this; so take a moment to analyze the context before jumping to conclusions about the score.