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The z-score is a measure of how many standard deviations a given value (in this case, infant length) is away from the mean. We can calculate the z-score using the formula: \[z = \frac{x - \mu}{\sigma}\] where \(z\) is the z-score, \(x\) is the infant length, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. a. For an infant length of 22 inches: \[z = \frac{22 - 20}{2.6} \approx 0.769\] b. For an infant length of 18 inches: \[z = \frac{18 - 20}{2.6} \approx -0.769\] c. For infant lengths between 19 and 21 inches, we'll need to calculate the z-scores for both 19 inches and 21 inches: \[z_{19} = \frac{19 - 20}{2.6} \approx -0.385\] \[z_{21} = \frac{21 - 20}{2.6} \approx 0.385\]

Now that we have the z-scores, we can use the standard normal distribution table (or a calculator) to find the corresponding probabilities. a. The probability of an infant having a length of more than 22 inches (z-score ≈ 0.769) is equal to the area under the curve to the right of the z-score, which is: \[\textbf{P}(z > 0.769) = 1 - \textbf{P}(z \leq 0.769) = 1 - 0.7794 \approx 0.2206\] b. The probability of an infant having a length of less than 18 inches (z-score ≈ -0.769) is equal to the area under the curve to the left of the z-score, which is: \[\textbf{P}(z < -0.769) = 0.2210\] c. The probability of an infant having a length between 19 and 21 inches (z-scores ≈ -0.385 and 0.385) is equal to the area under the curve between these two z-scores: \[\textbf{P}(-0.385 \leq z \leq 0.385) = \textbf{P}(z \leq 0.385) - \textbf{P}(z \leq -0.385) = 0.6501 - 0.3499 = 0.3002\] Solution: a. The probability that an infant has a length of more than 22 inches is approximately 0.2206 (or 22.06%). b. The probability that an infant has a length of less than 18 inches is approximately 0.2210 (or 22.10%). c. The probability that an infant has a length between 19 and 21 inches is approximately 0.3002 (or 30.02%).