Question

The profit (or loss) from an investment is normally distributed with a mean of $11,200 and a standard deviation of $8,250. What is the probability that there will be a loss rather than a profit? 0.15 \6.70% \0.93 \8.73%

Answer 1

The probability of experiencing a loss rather than a profit from the investment, given a normal distribution with a mean of $11,200 and a standard deviation of $8,250, is approximately 8.73%.

Step-by-step explanation:

In a normal distribution, probabilities are determined by calculating the area under the curve. To find the probability of a loss, we need to calculate the area to the left of zero (as a loss represents negative profits) on the normal distribution curve. Using the Z-score formula
`\(Z = \frac{{X - \mu}}{\sigma}\)` , where \(X\) is the value we're interested in (in this case, zero for a loss),
`\(\mu\)` is the mean, and
`\(\sigma\)` is the standard deviation, we find the Z-score for a loss:
`\(Z = \frac{{0 - 11,200}}{8,250} = -1.36\).`

By referring to a standard normal distribution table or using a calculator or statistical software, we can find the probability corresponding to this Z-score, which indicates the likelihood of observing a loss. This probability is approximately 0.0863 or 8.63%. Hence, there's an approximately 8.73% chance of experiencing a loss rather than a profit from this investment.

Understanding the probabilities in a normal distribution curve helps assess the likelihood of different outcomes. In this case, the negative Z-score signifies the probability of observing a loss, indicating that there's an approximately 8.73% chance that the investment will result in a loss rather than a profit, based on the given mean and standard deviation.