Explanation Approach: The bell-shaped normal curve is symmetric with respect to its mean (denoted as μ ). Thus, 50% of the data values lies to the left of mean and remaining 50% of the data lies to the right of the mean. Figure 1 illustrates 68-95-99.7 rule for a normal distribution . Figure 1 Approximately 68% of the data lies between μ − σ and μ + σ , 95% of the data lies between μ − 2 σ and μ + 2 σ and 99.7% of the data lies between μ − 3 σ and μ + 3 σ . Assume a normal distribution has a mean of μ and a standard deviation of σ . We use the equation z = x − μ σ to convert a value x in the nonstandard distribution to a z -score. Given Information: The mean of the distribution is 10 and the standard deviation is 2.2. Calculation: Substitute μ = 10 , σ = 2.2 and x = 16 in z = x − μ σ , z = 16 − 10 2.2 = 6 2.2 ≈ 2.73 Therefore, the corresponding z -score is 2.73. That is, P ( x ≥ 16 ) = P ( z ≥ 2.73 ) . Figure 1 To find the area between z = 0 and z = 2.73 , use the data in the standard normal distribution table. Use the row for z = 2.7 and the column labelled 0.03 in the standard normal distribution table. The data from the standard normal distribution table required for this problem is given below:

6th Edition

ISBN: 9780134434681

Author: Tom Pirnot

Publisher: PEARSON

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Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 14.4, Problem 81E

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The percentage of population who carry at least 16 genes out of 20 genes that could affect anxiety.

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Chapter 14.4, Problem 80E

Chapter 14.4, Problem 82E

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Chapter 14 Solutions

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The percentage of population who carry at least 16 genes out of 20 genes that could affect anxiety.

Solution Summary: The author illustrates the 68-95-99.7 rule for a normal distribution. The bell-shaped normal curve is symmetric with respect to its mean.

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To find: The percentage of population who carry at least 16 genes out of 20 genes that could affect anxiety.

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Chapter 14 Solutions

To find:

The percentage of population who carry at least 16 genes out of 20 genes that could affect anxiety.