To solve this problem, we first need to understand the properties of the normal distribution.

1. The mean of the sum of independent normal random variables is the sum of their means.

2. The variance of the sum of independent normal random variables is the sum of their variances.

Given that the mean annual profit is $26 million and the standard deviation is $6 million, for 7 years:

1. The mean of the total profit for 7 years would be 7 * $26 million = $182 million.

2. The variance of the total profit for 7 years would be 7 * ($6 million)^2 = $252 million^2. Therefore, the standard deviation would be the square root of the variance, which is approximately $15.87 million.

We are asked to find the probability that the total profit in the next 7 years is greater than $210 million.

To find this, we need to convert $210 million to a z-score. The z-score is calculated as (X - μ) / σ, where X is the value we are interested in, μ is the mean, and σ is the standard deviation.

So, the z-score for $210 million is ($210 million - $182 million) / $15.87 million ≈ 1.76.

We can then look up this z-score in a standard normal distribution table or use a calculator to find the probability. The table or calculator gives us the probability that the value is less than $210 million. But we want the probability that the value is greater than $210 million, so we subtract the value from 1.

The probability that a standard normal random variable is less than 1.76 is approximately 0.961. Therefore, the probability that it is greater than 1.76 is 1 - 0.961 = 0.039.

So, the probability that the total profit in the next 7 years is greater than $210 million is approximately 0.039, or 3.9%.

Therefore, the correct answer is 0.039.

Question

To solve this problem, we first need to understand the properties of the normal distribution.

1. The mean of the sum of independent normal random variables is the sum of their means.

2. The variance of the sum of independent normal random variables is the sum of their variances.

Given that the mean annual profit is $26 million and the standard deviation is $6 million, for 7 years:

1. The mean of the total profit for 7 years would be 7 * $26 million = $182 million.

2. The variance of the total profit for 7 years would be 7 * ($6 million)^2 = $252 million^2. Therefore, the standard deviation would be the square root of the variance, which is approximately $15.87 million.

We are asked to find the probability that the total profit in the next 7 years is greater than $210 million.

To find this, we need to convert $210 million to a z-score. The z-score is calculated as (X - μ) / σ, where X is the value we are interested in, μ is the mean, and σ is the standard deviation.

So, the z-score for $210 million is ($210 million - $182 million) / $15.87 million ≈ 1.76.

We can then look up this z-score in a standard normal distribution table or use a calculator to find the probability. The table or calculator gives us the probability that the value is less than $210 million. But we want the probability that the value is greater than $210 million, so we subtract the value from 1.

The probability that a standard normal random variable is less than 1.76 is approximately 0.961. Therefore, the probability that it is greater than 1.76 is 1 - 0.961 = 0.039.

So, the probability that the total profit in the next 7 years is greater than $210 million is approximately 0.039, or 3.9%.

Therefore, the correct answer is 0.039.

Knowee AI · Accepted Answer