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Chapter 6: Problem 29

Find the following areas under a normal distribution curve with \(\mu=20\) and \(\sigma=4\). a. Area between \(x=20\) and \(x=27\) b. Area from \(x=23\) to \(x=26\) c. Area between \(x=9.5\) and \(x=17\)

Short Answer

Expert verified
Area between \(x=20\) and \(x=27\) is approximately 0.4599. Area between \(x=23\) and \(x=26\) is approximately 0.1598. Area between \(x=9.5\) and \(x=17\) is approximately 0.2266.

Step by step solution

01

Calculate the z-scores for part (a)

First, calculate the z-scores for \(x=20\) and \(x=27\). By substituting in the values for \(x\), \(\mu\), and \(\sigma\), we find that the z-score for \(x=20\) is \(z=(20-20)/4 = 0\) and for \(x=27\) is \(z=(27-20)/4 = 1.75\).
02

Look up the z-scores in the Z table for part (a)

The value of a standard normal distribution at \( z = 0 \) is always \( 0.5 \). For \( z = 1.75 \), look up this value in the Z table. The area to the left of \( z = 1.75 \) under the normal distribution is about \( 0.9599 \). The required area between \(x=20\) and \(x=27\) is therefore \( 0.9599 - 0.5 = 0.4599 \).
03

Calculate the z-scores for part (b)

Here, calculate the z-scores for \(x=23\) and \(x=26\). Following the same process as before, we find that the z-score for \(x=23\) is \(z=(23-20)/4 = 0.75\) and for \(x=26\) is \(z=(26-20)/4 = 1.5\).
04

Look up the z-scores in the Z table for part (b)

Next, we look these z-scores up in the Z table. The area to the left of \( z = 0.75 \) under the normal distribution is about \( 0.7734 \), and for \( z = 1.5 \), it is about \( 0.9332 \). Therefore, the required area between \(x=23\) and \(x=26\) is \( 0.9332 - 0.7734 = 0.1598 \).
05

Calculate the z-scores for part (c)

Finally, calculate the z-scores for \(x=9.5\) and \(x=17\). Here, the z-score for \(x=9.5\) is \(z=(9.5-20)/4 = -2.625\) and for \(x=17\) is \(z=(17-20)/4 = -0.75\).
06

Look up the z-scores in the Z table for part (c)

The area to the left of \( z = -2.625 \) under the normal distribution is technically very minimal, and for \( z = -0.75 \), it is \( 0.2266 \). Thus, the required area between \(x=9.5\) and \(x=17\) is \( 0.2266 - 0 = 0.2266 \).

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