4. The average family income in the town of Seneville is $45 000, with a standard
deviation of $9000. Assuming that family income is normally distributed,
determine the income level below which 80% of the families in this town earn
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.8000) and its Z score. Insert Z score in the equation above and solve for score.
To determine the income level below which 80% of the families in the town of Seneville earn, we can use the z-score formula and the standard normal distribution.
1. First, we need to find the z-score corresponding to the given percentile (80%). The z-score represents the number of standard deviations a value is away from the mean.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
- x is the income level that we want to find
- μ is the mean family income (45,000)
- σ is the standard deviation (9,000)
2. To find the z-score corresponding to the 80th percentile, we need to look up the z-score in the standard normal distribution table. The table provides the area under the curve for different z-scores.
Looking up the z-score in the table, we find that the z-score for the 80th percentile is approximately 0.84.
3. Now we can rearrange the z-score formula to solve for x (the income level we want to find):
x = z * σ + μ
Plugging in the values:
x = 0.84 * 9,000 + 45,000
Calculating this, we get:
x = 7,560 + 45,000
Therefore, the income level below which 80% of the families in Seneville earn is approximately $52,560.