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Determine the total area under the standard normal curve in parts (a) through (c) below. (a) Find the area under the normal curve to the left of \( z=-1 \) plus the area under the normal curve to the right of \( z=1 \). The combined area is \( \square \). (Round to four decimal places as needed.) (b) Find the area under the normal curve to the left of \( z=-1.55 \) plus the area under the normal curve to the right of \( z=2.55 \). The combined area is \( \square \). (Round to four decimal places as needed.) (c) Find the area under the normal curve to the left of \( z=-0.23 \) plus the area under the normal curve to the right of \( z=1.10 \). The combined area is \( \square \). (Round to four decimal places as needed.)

Ask by Gardner Page. in the United States
Mar 15,2025

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Answer

(a) 0.3174 (b) 0.0660 (c) 0.5447

Solution

(a) The area to the left of \( z=-1 \) is \[ P(z < -1) = 1 - \Phi(1) \,, \] and the area to the right of \( z=1 \) is \[ P(z > 1) = 1 - \Phi(1) \,. \] Since the standard normal curve is symmetric, we have \[ P(z < -1) + P(z > 1) = 2\left(1 - \Phi(1)\right)\,. \] Using \(\Phi(1) \approx 0.8413\), \[ 2(1 - 0.8413) = 2(0.1587) = 0.3174\,. \] (b) The area to the left of \( z=-1.55 \) is \[ P(z < -1.55) = 1 - \Phi(1.55)\,. \] For \( z=1.55 \), \(\Phi(1.55) \approx 0.9394\), so \[ P(z < -1.55) \approx 1 - 0.9394 = 0.0606\,. \] The area to the right of \( z=2.55 \) is \[ P(z > 2.55) = 1 - \Phi(2.55)\,. \] For \( z=2.55 \), \(\Phi(2.55) \approx 0.9946\), so \[ P(z > 2.55) \approx 1 - 0.9946 = 0.0054\,. \] The combined area is \[ 0.0606 + 0.0054 = 0.0660\,. \] (c) The area to the left of \( z=-0.23 \) is \[ P(z < -0.23) = 1 - \Phi(0.23)\,. \] For \( z=0.23 \), \(\Phi(0.23) \approx 0.5910\), so \[ P(z < -0.23) \approx 1 - 0.5910 = 0.4090\,. \] The area to the right of \( z=1.10 \) is \[ P(z > 1.10) = 1 - \Phi(1.10)\,. \] For \( z=1.10 \), \(\Phi(1.10) \approx 0.8643\), so \[ P(z > 1.10) \approx 1 - 0.8643 = 0.1357\,. \] The combined area is \[ 0.4090 + 0.1357 = 0.5447\,. \] Final answers: (a) \( 0.3174 \) (b) \( 0.0660 \) (c) \( 0.5447 \)

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To find the total area under the standard normal curve for each part, we utilize the properties of the standard normal distribution, where the total area under the curve equals 1. The values of \( z \) correspond to cumulative probabilities in standard normal tables or calculators. For part (a): The area to the left of \( z=-1 \) is approximately 0.1587, and the area to the right of \( z=1 \) is also about 0.1587. Therefore, the combined area is: \[ 0.1587 + 0.1587 = 0.3174 \] For part (b): The area to the left of \( z=-1.55 \) is approximately 0.0606, and the area to the right of \( z=2.55 \) is about 0.0054. Thus, the combined area is: \[ 0.0606 + 0.0054 = 0.0660 \] For part (c): The area to the left of \( z=-0.23 \) is approximately 0.4099, and the area to the right of \( z=1.10 \) is about 0.1359. Hence, the combined area is: \[ 0.4099 + 0.1359 = 0.5458 \] So, the combined areas respectively are: (a) 0.3174 (b) 0.0660 (c) 0.5458 Now, let’s spice up the fun with some engaging insights! The world of standard normal distribution is like a massive party where everyone knows the rules—every guest (think of every z-value!) has a certain area of trust on either side of them. You can always count on those z-scores to show you where the probabilities hang out, making your statistical adventure much smoother and more predictable! If you’re attending a statistics convention—and we’re hoping you do—remember that keeping a z-table handy or having access to online calculators can be a total game-changer! It’s like having a GPS for navigating the sometimes confusing terrain of probability and area calculations. Don’t forget to practice; it’s the best way to shine bright like a star in the normal distribution sky!

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