The average income tax refund for the 2009 tax year was 52876_ Assume the refund per person follows the norma probability distribution with standard deviation of 5971 . Complete parts through below
The Correct Answer and Explanation is:
To solve this problem, we are working with a normal distribution. The average (mean) income tax refund for the 2009 tax year is $5,287.60, and the standard deviation is $597.10. We’ll interpret the parts you are referring to as likely involving calculations with the normal distribution, such as finding probabilities or values corresponding to certain percentiles.
Let’s walk through three typical parts that often come with a normal distribution problem:
a) Find the probability that a randomly selected person received a refund greater than $6,000.
We use the z-score formula:z=X−μσz = \frac{X – \mu}{\sigma}z=σX−μ
Where:
- X=6000X = 6000X=6000
- μ=5287.60\mu = 5287.60μ=5287.60
- σ=597.10\sigma = 597.10σ=597.10
z=6000−5287.60597.10≈712.40597.10≈1.193z = \frac{6000 – 5287.60}{597.10} \approx \frac{712.40}{597.10} \approx 1.193z=597.106000−5287.60≈597.10712.40≈1.193
Now, we look up z=1.19z = 1.19z=1.19 on the standard normal distribution table (or use a calculator). The area to the left of z=1.19z = 1.19z=1.19 is about 0.8830.
So, the probability of receiving more than $6,000 is:P(X>6000)=1−P(Z<1.19)=1−0.8830=0.1170P(X > 6000) = 1 – P(Z < 1.19) = 1 – 0.8830 = 0.1170P(X>6000)=1−P(Z<1.19)=1−0.8830=0.1170
Answer: 0.1170 or 11.70%
b) Find the probability that a refund is less than $5,000.
z=5000−5287.60597.10=−287.60597.10≈−0.4817z = \frac{5000 – 5287.60}{597.10} = \frac{-287.60}{597.10} \approx -0.4817z=597.105000−5287.60=597.10−287.60≈−0.4817
Using the z-table, the area to the left of z=−0.48z = -0.48z=−0.48 is about 0.3156.
Answer: 0.3156 or 31.56%
c) What refund amount represents the top 10% of refunds?
We are looking for XXX such that P(X>x)=0.10P(X > x) = 0.10P(X>x)=0.10, or P(Z<z)=0.90P(Z < z) = 0.90P(Z<z)=0.90. From the z-table, the z-value for the 90th percentile is approximately 1.28.
Now solve for XXX:X=μ+zσ=5287.60+1.28(597.10)≈5287.60+763.29=6050.89X = \mu + z\sigma = 5287.60 + 1.28(597.10) \approx 5287.60 + 763.29 = 6050.89X=μ+zσ=5287.60+1.28(597.10)≈5287.60+763.29=6050.89
Answer: Approximately $6,050.89
Explanation (300 words)
This problem is a classic example of applying the normal distribution, a core concept in statistics used to model data that clusters around a mean. The normal distribution is symmetric and defined by two parameters: the mean (average) and the standard deviation (a measure of spread).
The mean refund is $5,287.60, which represents the central point of the distribution. The standard deviation of $597.10 indicates how much variation exists from the mean. Larger values indicate more variability in refund amounts.
To find the probability that a person receives more than a specific amount (like $6,000), we compute the z-score, which measures how many standard deviations a value is from the mean. This score is then used to look up the corresponding probability from the standard normal distribution table. Subtracting this probability from 1 gives the proportion of people with refunds above that amount.
Similarly, to find the chance of receiving less than a certain amount, we use the same formula and interpret the area under the curve to the left of the z-score.
To identify refund amounts for the top 10%, we reverse the process: find the z-score that corresponds to the 90th percentile (since only 10% of refunds are higher), then convert it back into a dollar amount using the mean and standard deviation. This technique helps policymakers and analysts understand income patterns and prepare for future tax seasons.
These tools are widely used in fields like finance, healthcare, and education to make data-driven decisions.
