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Chapter 6: Problem 24

Find the following probabilities for the standard normal distribution. a. \(P(z<-1.31)\) b. \(P(1.23 \leq z \leq 2.89)\) c. \(P(-2.24 \leq z \leq-1.19)\) d. \(P(z<2.02)\)

Short Answer

Expert verified
a. 0.0950, b. 0.1074, c. 0.1045, d. 0.9783

Step by step solution

01

Probability for \(z

The z-score here is -1.31, which falls below the mean. The table gives the area to the right of the z-score but we want the area to the left, which is \(P(z<-1.31)\). Since the total probability under the curve is 1, we can find the required probability by subtracting the table value from 1. From the table, the area to the right of -1.31 is approximately 0.9050, so the required probability is \(1 - 0.9050 = 0.0950\).
02

Probability for \(1.23 \leq z \leq 2.89\)

This is a range of z-scores. To find this probability, find the areas under the curve up to each z-score from the table and subtract the smaller from the larger. The area to the left of 1.23 is approximately 0.8907, and the area to the left of 2.89 is approximately 0.9981, so the required probability is \(0.9981 - 0.8907 = 0.1074\).
03

Probability for \(-2.24 \leq z \leq -1.19\)

Both z-scores fall below the mean. Like in Step 2, look up the area to the left of each z-score and subtract the smaller from the larger. The area to the right of -2.24 is approximately 0.9875 (or 1 - 0.0125), and the area to the right of -1.19 is approximately 0.8830 (or 1 - 0.1170), so the required probability is \(0.9875 - 0.8830 = 0.1045\).
04

Probability for \(z

The z-score here is greater than the mean, so we can directly find the area to the left of the z-score as the required probability. From the table, the area to the left of 2.02 is approximately 0.9783, so \(P(z<2.02) = 0.9783\).

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