Suppose that the distribution of household incomes in the United States is approximately normal with a mean of 69.8 thousand dollars, and a standard deviation of 15 thousand dollars.
a) Find the probability that a randomly selected household would have an income between 66 and 73 thousand dollars.
To find the probability that a randomly selected household would have an income between 66 and 73 thousand dollars, we can use the standard normal distribution.
First, we need to convert the given values (66 and 73) into Z-scores. The Z-score is a measure of how many standard deviations an observation or value is from the mean of a distribution.
The Z-score is calculated using the formula:
Z = (X - μ) / σ
Where:
X is the given value (66 or 73),
μ is the mean of the distribution (69.8 thousand dollars), and
σ is the standard deviation of the distribution (15 thousand dollars).
For 66 thousand dollars:
Z1 = (66 - 69.8) / 15
For 73 thousand dollars:
Z2 = (73 - 69.8) / 15
Now, we can look up the Z-scores in the standard normal distribution table or use a statistical calculator to find the area under the curve between the two Z-scores. This area represents the probability.
Alternatively, we can use the Z-score formula to calculate the probabilities:
P(Z1 < Z < Z2) = P(Z < Z2) - P(Z < Z1)
Using the Z-score formula, we can calculate the corresponding probabilities.
After calculating the probabilities using the Z-score formula or looking them up in a standard normal distribution table, we will have the answer to the question.