Explanation Approach: 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 yearly return is 5.9% and the standard deviation is 1.8%. Calculation: Substitute μ = 5.9 % , σ = 1.8 % and x = 8 % in z = x − μ σ , z = 8 % − 5.9 % 1.8 % = 2.1 % 1.8 % ≈ 1.17 Therefore, the corresponding z -score is approximately 1.17. To find the area between z = 0 and to a z -score, use the data in the standard normal distribution table. Use the row for z = 1.1 and the column labelled 0.07 in the standard normal distribution table. The data from the standard normal distribution table required for this problem is given below: z value 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 To determine b) To find: The number of years required to earn less than 4% on the investment.

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 74E

To determine

a)

To find:

The number of years required to earn at least 8% on the investment.

To determine

b)

To find:

The number of years required to earn less than 4% on the investment.

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Chapter 14.4, Problem 73E

Chapter 14.4, Problem 75E

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a) To find: The number of years required to earn at least 8% on the investment.

Solution Summary: The author explains how to find the number of years required to earn at least 8% on the investment.