Respuesta :
Answer:
52.84% probability the itemized deductions were within $200 of the sample mean
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
[tex]\mu = 16642, \sigma = 2400, n = 75, s = \frac{2400}{\sqrt{75}} = 277.13[/tex]
What is the probability the itemized deductions were within $200 of the sample mean?
This is the pvalue of Z when X = 16642 + 200 = 16842 subtracted by the pvalue of Z when X = 16642 - 200 = 16442. So
X = 16842
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16842 - 16642}{277.13}[/tex]
[tex]Z = 0.72[/tex]
[tex]Z = 0.72[/tex] has a pvalue of 0.7642
X = 16442
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16442 - 16642}{277.13}[/tex]
[tex]Z = -0.72[/tex]
[tex]Z = -0.72[/tex] has a pvalue of 0.2358
0.7642 - 0.2358 = 0.5284
52.84% probability the itemized deductions were within $200 of the sample mean
