Let's break down each part of the problem and solve them using properties of the normal distribution.

The given information:

• The real income in Connecticut is normally distributed.
• Mean $\mu = 60,000$ (assuming this value based on typical data patterns; adjust as necessary).
• Standard deviation $\sigma = 12,000$.

a. Probability that an individual has an annual real income of $75,000$ or more.

To solve this, we use the z-score formula: $Z = \frac{X - \mu}{\sigma}$ Where:

• $X$ is the value we are interested in (here, $X = 75,000$).
• $\mu = 60,000$ (mean).
• $\sigma = 12,000$ (standard deviation).

So, the z-score for $75,000$ is: $Z = \frac{75,000 - 60,000}{12,000} = \frac{15,000}{12,000} = 1.25$

Using the z-score of 1.25, we can look up the corresponding probability from a standard normal distribution table or use a calculator.

• The cumulative probability $P(Z \leq 1.25)$ is approximately $0.8944$.

So, the probability that an individual has an income of $75,000$ or more is: $P(X \geq 75,000) = 1 - P(Z \leq 1.25) = 1 - 0.8944 = 0.1056$

Thus, the probability is approximately 0.11 (rounded to 2 decimal places).

b. Probability that an individual has an annual real income of $50,000$ or less.

First, calculate the z-score for $50,000$: $Z = \frac{50,000 - 60,000}{12,000} = \frac{-10,000}{12,000} = -0.8333$

Using the z-score of -0.83, the cumulative probability $P(Z \leq -0.83)$ is approximately $0.2033$.

Thus, the probability that an individual has an income of $50,000$ or less is about 0.2033, or 20% (rounded to 0 decimal place, this would be 0).

c. Probability that an individual has an annual real income between $55,000$ and $65,000$.

We need to find the z-scores for both $55,000$ and $65,000$:

1. For $X = 55,000$: $Z = \frac{55,000 - 60,000}{12,000} = \frac{-5,000}{12,000} = -0.4167$ The cumulative probability $P(Z \leq -0.4167)$ is approximately $0.3385$.

2. For $X = 65,000$: $Z = \frac{65,000 - 60,000}{12,000} = \frac{5,000}{12,000} = 0.4167$ The cumulative probability $P(Z \leq 0.4167)$ is approximately $0.6615$.

To find the probability that an individual has an income between $55,000$ and $65,000$: $P(55,000 \leq X \leq 65,000) = P(Z \leq 0.4167) - P(Z \leq -0.4167) = 0.6615 - 0.3385 = 0.323$

Thus, the probability is approximately 0.32 (rounded to 2 decimal places).

d. Annual real income at the 90th percentile.

To find the income corresponding to the 90th percentile, we first look up the z-score corresponding to a cumulative probability of 0.90. The z-score is approximately $1.28$.

We now use the z-score formula in reverse to find the income: $X = \mu + Z \cdot \sigma = 60,000 + 1.28 \cdot 12,000 = 60,000 + 15,360 = 75,360$

Thus, the annual real income at the 90th percentile is approximately 75,360 (rounded to 2 decimal places).

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Would you like further details on any part of the solution?

Here are 5 additional questions you could consider:

1. How would the probabilities change if the standard deviation was higher or lower?
2. What is the probability that an individual earns more than $85,000$?
3. How would the 90th percentile change if the mean income increased to $70,000$?
4. What is the income level at the 50th percentile (median)?
5. How does real income per capita account for inflation when compared to nominal income?

Tip: Z-scores help translate normal distribution data into a standardized form, making it easier to find probabilities and percentiles.