Math  /  Data & Statistics

QuestionQuestion Find the area to the left of the zz-score -1.35 under the standard normal curve. \begin{tabular}{c|ccccccccccc} z & 0.00\mathbf{0 . 0 0} & 0.01\mathbf{0 . 0 1} & 0.02\mathbf{0 . 0 2} & 0.03\mathbf{0 . 0 3} & 0.04\mathbf{0 . 0 4} & 0.05\mathbf{0 . 0 5} & 0.06\mathbf{0 . 0 6} & 0.07\mathbf{0 . 0 7} & 0.08\mathbf{0 . 0 8} & 0.09\mathbf{0 . 0 9} \\ \hline 1.5\mathbf{- 1 . 5} & 0.0668 & 0.0655 & 0.0643 & 0.0630 & 0.0618 & 0.0606 & 0.0594 & 0.0582 & 0.0571 & 0.0559 \\ 1.4-\mathbf{1 . 4} & 0.0808 & 0.0793 & 0.0778 & 0.0764 & 0.0749 & 0.0735 & 0.0721 & 0.0708 & 0.0694 & 0.0681 \\ 1.3\mathbf{- 1 . 3} & 0.0968 & 0.0951 & 0.0934 & 0.0918 & 0.0901 & 0.0885 & 0.0869 & 0.0853 & 0.0838 & 0.0823 \\ 1.2\mathbf{- 1 . 2} & 0.1151 & 0.1131 & 0.1112 & 0.1093 & 0.1075 & 0.1056 & 0.1038 & 0.1020 & 0.1003 & 0.0985 \\ 1.1\mathbf{- 1 . 1} & 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 1.3-1.3 and the column 0.050.05 in the table. The value is 0.08850.0885.
The area to the left of the z-score 1.35-1.35 under the standard normal curve is:
0.0885 \boxed{0.0885}

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