Question

imagine that chris’s score is 3 and the mean of the class is 30. you have not seen the distribution of raw scores but are told that the grade distribution is approximately normal and that his score is 1 standard deviation below the mean. calculate the standard deviation of the class grades to two decimal places, assuming that you are using this value to estimate the standard deviation of grades across the school. these are the following equations for the mean, standard deviation, and z-score: (\frac{sum x}{n}), (sqrt{\frac{(sum (x - \bar{x})^2}{n - 1}}), (z = \frac{x - \bar{x}}{sigma}) (\bigcirc) 15 (\bigcirc) 5 (\bigcirc) 20 (\bigcirc) 27

Explanation:

Step1: Identify given values

We know that Chris's score \( X = 3 \), the mean \( \bar{X}=30 \), and the z - score \( Z=- 1 \) (since his score is 1 standard deviation below the mean). The formula for the z - score is \( Z=\frac{X - \bar{X}}{\sigma} \).

Step2: Rearrange the z - score formula to solve for \( \sigma \)

From \( Z=\frac{X - \bar{X}}{\sigma} \), we can rearrange it to \( \sigma=\frac{X - \bar{X}}{Z} \).

Step3: Substitute the values into the formula

Substitute \( X = 3 \), \( \bar{X}=30 \), and \( Z=-1 \) into the formula:
\( \sigma=\frac{3 - 30}{-1}=\frac{- 27}{-1}=27.00 \)

Answer:

27 (or 27.00 when considering two decimal places)