d) What is the probability a person in this sample have a cholesterol level belo

Question

d) What is the probability a person in this sample have a cholesterol level below 170 mg/di? Problem 25, page 316: The Internal Revenue Service reports that the mean federal income tax paid in the year 2010 was $8040. Assume that the standard deviation is $5000. The IRS plans to draw a sample of 1000 tax returns to study the effects of a new tax law. a) What is the probability the mean tax of this sample is less than $8000? b) What is the probability that the mean tax of the sample is between $7600 and $7900? c) Find the 40th percentile of the sample mean of taxes? d) Would it be unusual if the sample mean were less than $7500? e) Would it be unusual if an individual in this sample paid less than $7500 in taxes. Why?

Explanation / Answer

We are given

Population mean = µ = 8040

Population standard deviation = = 5000

Sample size = n = 1000

Formula for Z score is given as below:

Z = (Xbar - µ) / [/sqrt(1000)]

Part a

Here, we have to find P(Xbar<8000)

Z = (8000 – 8040) / [5000/sqrt(1000)]

Z = 40/ 158.1139

Z = 0.252982

P(Z< 0.252982) = P(Xbar<8000) = 0.599859

(by using normal table or excel)

Required probability = 0.599859

Part b

We have to find P(7600<Xbar<7900)

P(7600<Xbar<7900) = P(Xbar<7900) – P(Xbar<7600)

For Xbar<7900

Z = (7900 – 8040) / [5000/sqrt(1000)]

Z = -0.88544

P(Z<-0.88544) = P(Xbar<7900) = 0.18796

(by using normal table or excel)

For Xbar<7600

Z = (7600 – 8040) / [5000/sqrt(1000)]

Z = -2.7828

P(Z<-2.7828) = P(Xbar<7900) = 0.002695

(by using normal table or excel)

P(7600<Xbar<7900) = P(Xbar<7900) – P(Xbar<7600)

P(7600<Xbar<7900) = 0.599859 - 0.002695

P(7600<Xbar<7900) = 0.597164

Required probability = 0.597164

Part c

We have to find 40th percentile of taxes.

Xbar = µ + Z*

Z = -0.25335

Xbar = 8040 + (-0.25335)*5000

Xbar = 6773.25

Required answer = 6773.25

Part d

We have to find P(Xbar<7500)

Z = (7500 – 8040) / [5000/sqrt(1000)]

Z = -540/158.1139

Z = -3.41526

P(Z<-3.41526) = 0.000319

Yes, it is unusual if the sample mean were less than $7500.

(Probability is unusual if it is less than 0.05)

Part e

Here, we have to consider single individual only.

Z = (X – µ) /

We have to find P(X<7500)

Z = (7500 – 8040) / 5000

Z = -0.108

P(Z<-0.108) = 0.456998

No, it is not unusual if an individual in this sample paid less than $7500 in taxes, because we get probability greater than 0.05.

[All probabilities are calculated by using normal table or excel.]

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