QuestionQuestion
Find the area to the left of the -score -1.35 under the standard normal curve.
\begin{tabular}{c|ccccccccccc}
z & & & & & & & & & & \\
\hline & 0.0668 & 0.0655 & 0.0643 & 0.0630 & 0.0618 & 0.0606 & 0.0594 & 0.0582 & 0.0571 & 0.0559 \\
& 0.0808 & 0.0793 & 0.0778 & 0.0764 & 0.0749 & 0.0735 & 0.0721 & 0.0708 & 0.0694 & 0.0681 \\
& 0.0968 & 0.0951 & 0.0934 & 0.0918 & 0.0901 & 0.0885 & 0.0869 & 0.0853 & 0.0838 & 0.0823 \\
& 0.1151 & 0.1131 & 0.1112 & 0.1093 & 0.1075 & 0.1056 & 0.1038 & 0.1020 & 0.1003 & 0.0985 \\
& 0.1357 & 0.1335 & 0.1314 & 0.1292 & 0.1271 & 0.1251 & 0.1230 & 0.1210 & 0.1190 & 0.1170
\end{tabular}
Use the value(s) from the table above.
Provide your answer below:
Studdy Solution
Locate the intersection of the row and the column in the table. The value is .
The area to the left of the z-score under the standard normal curve is:
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