Use the Partial Table Below to Find the P-Value That $\underline{\text { Standard Score and Percentiles for a Normal Distribution }}$

Use the partial table below to find the P-value that corresponds to the standard z-score, and determine whether the
alternative hypothesis is supported or not supported at the 0.05 significance level. $\underline{\text { Standard Score and Percentiles for a Normal Distribution }}$
$\begin{array}{rlrlll}\underline{Z }&\underline{ \% }\ &\underline{ Z }& \underline{\% }&\underline{Z }& \underline{\%}\\ -3.5 & 0.02 & -1.0 & 15.87 & 1.5 & 93.32 \\-3.0 & 0.13 & -0.5 & 30.85 & 2.0 & 97.72 \\-2.5 & 0.62 & 0.0 & 50.00 & 2.5 & 99.38 \\-2.0 & 2.28 & 0.5 & 69.15 & 3.0 & 99.87 \\-1.5 & 6.68 & 1.0 & 84.13 & 3.5 & 99.98\end{array}$

- $$\mathrm { z } = - 0.5 \text { for } \mathrm { H } _ { \mathrm { a } } : \mu < 60$$

A) 0.3085; supported
B) $0.3085$ ; not supported
C) $0.6915$ ; not supported
D) $0.6915$ ; supported