Question Determine the area under the standard normal curve that lies to the right of the $z$-score -1.12 and to the left of the $z$-score -0.79 . \begin{tabular}{c|cccccccccc} $\mathbf{z}$ & $\mathbf{0 . 0 0}$ & $\mathbf{0 . 0 1}$ & $\mathbf{0 . 0 2}$ & $\mathbf{0 . 0 3}$ & $\mathbf{0 . 0 4}$ & $\mathbf{0 . 0 5}$ & $\mathbf{0 . 0 6}$ & $\mathbf{0 . 0 7}$ & $\mathbf{0 . 0 8}$ & $\mathbf{0 . 0 9}$ \\ \hline $\mathbf{- 1 . 2}$ & $\mathbf{0 . 1 1 5 1}$ & $\mathbf{0 . 1 1 3 1}$ & $\mathbf{0 . 1 1 1 2}$ & 0.1093 & $\mathbf{0 . 1 0 7 5}$ & 0.1056 & 0.1038 & 0.1020 & 0.1003 & 0.0985 \\ $\mathbf{- 1 . 1}$ & 0.1357 & 0.1335 & 0.1314 & 0.1292 & 0.1271 & 0.1251 & 0.1230 & 0.1210 & 0.1190 & 0.1170 \\ $\mathbf{- 1 . 0}$ & 0.1587 & 0.1562 & 0.1539 & 0.1515 & 0.1492 & 0.1469 & 0.1446 & 0.1423 & 0.1401 & 0.1379 \\ $\mathbf{- 0 . 9}$ & 0.1841 & 0.1814 & 0.1788 & 0.1762 & 0.1736 & 0.1711 & 0.1685 & 0.1660 & 0.1635 & 0.1611 \\ $\mathbf{- 0 . 8}$ & 0.2119 & 0.2090 & 0.2061 & 0.2033 & 0.2005 & 0.1977 & 0.1949 & 0.1922 & 0.1894 & 0.1867 \\ $\mathbf{- 0 . 7}$ & 0.2420 & 0.2389 & 0.2358 & 0.2327 & 0.2296 & 0.2266 & 0.2236 & 0.2206 & 0.2177 & 0.2148 \\ $\mathbf{- 0 . 6}$ & 0.2743 & 0.2709 & 0.2676 & 0.2643 & 0.2611 & 0.2578 & 0.2546 & 0.2514 & 0.2483 & 0.2451 \end{tabular} Use the value(s) from the table above. Provide your answer below:
Solution
Step 1 ：The question is asking for the area under the standard normal curve between two z-scores. The area under the curve for a standard normal distribution represents the probability of a random variable falling within a certain range. In this case, we want to find the probability that a random variable falls between z-scores of -1.12 and -0.79.
Step 2 ：To find this, we need to find the area to the left of z = -0.79 and subtract the area to the left of z = -1.12. This is because the area to the left of z = -0.79 includes the area we want, but also includes the area to the left of z = -1.12, which we don't want. So, we subtract the area to the left of z = -1.12 to get the area between the two z-scores.
Step 3 ：From the table, we can see that the area to the left of z = -0.79 is 0.2148 and the area to the left of z = -1.12 is 0.1314.
Step 4 ：Subtract the area to the left of z = -1.12 from the area to the left of z = -0.79 to get the area between the two z-scores.
Step 5 ：\(z\_score\_left = 0.2148\)
Step 6 ：\(z\_score\_right = 0.1314\)
Step 7 ：\(area\_between = z\_score\_left - z\_score\_right\)
Step 8 ：\(area\_between = 0.0834\)
Step 9 ：Final Answer: The area under the standard normal curve that lies to the right of the z-score -1.12 and to the left of the z-score -0.79 is \(\boxed{0.0834}\).
