Question

It is well documented that a typical washing machine can last anywhere between 5 to 20 years. Let the life of a washing machine be represented by a lognormal variable, Y = eX where X is normally distributed. In addition, let the mean and standard deviation of the life of a washing machine be 5 and half years and 3 years, respectively. [You may find it useful to reference the z table.]

a. Compute the mean and the standard deviation of X. (Round your intermediate calculations to at least 4 decimal places and final answers to 4 decimal places.)



b. What proportion of the washing machines will last for more than 11 years? (Round your intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)



c. What proportion of the washing machines will last for less than 3 years? (Round your intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)



d. Compute the 70th percentile of the life of the washing machines. (Round your intermediate calculations to at least 4 decimal places and final answer to the nearest whole number.)

Answer 1

a. The mean and standard deviation of X are:

mean(X) = ln(5.5) - (1/2) * ln(1 + (3/5.5)^2) ≈ 1.5041 years

std(X) = sqrt(ln(1 + (3/5.5)^2)) ≈ 0.8321 years

b. Let Z be the standard normal variable corresponding to the life of 11 years. Then:

Z = (ln(11) - ln(5.5)) / 0.8321 ≈ 0.9672

Using the z-table, we find that the proportion of the washing machines that will last for more than 11 years is:

P(Z > 0.9672) ≈ 0.1661

c. Let Z be the standard normal variable corresponding to the life of 3 years. Then:

Z = (ln(3) - ln(5.5)) / 0.8321 ≈ -1.4645

Using the z-table, we find that the proportion of the washing machines that will last for less than 3 years is:

P(Z < -1.4645) ≈ 0.0721

d. The 70th percentile of the life of the washing machines corresponds to the standard normal variable Z such that:

P(Z < z) = 0.70

Using the z-table, we find that z ≈ 0.5244. Then:

ln(Y) = mean(X) + z * std(X) ≈ 2.1629

Solving for Y, we get:

Y ≈ e^2.1629 ≈ 8.6913 years

Therefore, the 70th percentile of the life of the washing machines is about 9 years.

Answer 2

a. The mean of X is μ = ln(5.5) - 0.5 * (ln(3/5.5))^2 ≈ -0.8329. The standard deviation of X is σ = sqrt(ln(3/5.5)^2) ≈ 0.7634.

b. We want to find P(Y > 11) = P(e^X > 11) = P(X > ln(11)) = P(Z > (ln(11) - ln(5.5) + 0.5 * ln(3/5.5)^2) / 0.7634) ≈ 0.0002, where Z is the standard normal distribution.

c. We want to find P(Y < 3) = P(e^X < 3) = P(X < ln(3)) = P(Z < (ln(3) - ln(5.5) + 0.5 * ln(3/5.5)^2) / 0.7634) ≈ 0.0001, where Z is the standard normal distribution.

d. The 70th percentile corresponds to a z-score of z = invNorm(0.7) ≈ 0.5244. Therefore, we want to find the value of y such that P(Y < y) = P(e^X < y) = 0.7. This implies that ln(y) = ln(5.5) - 0.8329 * 0.7634 + 0.5244 * 0.7634, so y ≈ 7.