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Find the z-score such that the area under the standard normal curve to the left is 0.19 . is the z-score such that the area under the curve to the left is 0.19 . (Round to two decimal places as needed.)

Ask by Summers Lee. in the United States
Mar 17,2025

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Answer

The \( z \)-score is approximately -0.88.

Solution

1. We need to find the value of \( z \) such that \[ P(Z < z) = 0.19. \] 2. Since \( 0.19 < 0.5 \), \( z \) must be negative. We use the inverse cumulative distribution (or a \( z \)-table) for the standard normal distribution to find the appropriate value of \( z \). 3. Looking up the cumulative probability closest to 0.19 in a \( z \)-table, we find that: \[ P(Z < -0.88) \approx 0.1894. \] This value is very close to 0.19. 4. Rounding \( z \) to two decimal places, we have: \[ z \approx -0.88. \] Therefore, the \( z \)-score such that the area under the standard normal curve to the left is 0.19 is \( z \approx -0.88 \).

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To find the z-score corresponding to an area of 0.19 to the left under the standard normal curve, you can use a standard normal distribution table or a calculator. Looking up the value, you find that the z-score is approximately -0.88 when rounded to two decimal places. This means that about 19% of the data falls below this z-score on the standard normal distribution. When working with z-scores and normal distributions, a common mistake is to misinterpret the area as the z-score or to round prematurely. Always ensure to look up the precise values and use a reliable source or tool to avoid errors in this crucial step!

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