Submit test A test score of 48.4 on a test having a mean of 66 and a standard deviation of 11 . Find the $z$-score corresponding to the given value and use the $z$-score to determine whether the value is significant. Consider a score to be significant if its z-score is less than -2.00 or greater than 2.00 . Round the $z$-score to the nearest tenth if necessary. A. -17.6 ; significant B. -1.6 , significant C. 1.6 ; not significant D. -1.6 ; not significant Search

Step 1 ：Given a test score of 48.4, a mean of 66, and a standard deviation of 11, we are asked to find the $z$-score and determine whether the value is significant. A score is considered significant if its $z$-score is less than -2.00 or greater than 2.00.

Step 2 ：The $z$-score is a measure of how many standard deviations an element is from the mean. The formula to find the $z$-score is: \(z = \frac{X - \mu}{\sigma}\), where \(z\) is the $z$-score, \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 ：Substituting the given values into the formula, we get: \(z = \frac{48.4 - 66}{11}\).

Step 4 ：Solving the above expression, we find that the $z$-score is -1.6.

Step 5 ：Since the $z$-score is not less than -2.00 or greater than 2.00, the score is not significant.

Step 6 ：Final Answer: The $z$-score is \(\boxed{-1.6}\) and the score is \textbf{not significant}.