a) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?
b) What is the probability that the mean of a sample of 50 tax returns is more than $289?
Solution :
Given that ,
mean = = 273
standard deviation = = 100
a ) n = 30
= = 273
= / n = 100/ 30 = 18.2574
within 16 = 273 ± 16 = 257, 289
P(257 < < 138)
= P[(257 - 273) / 18.2574< ( - ) / < (289 - 273) /18.2574 )]
= P(-0.88 < Z < 0.88)
= P(Z < 0.88) - P(Z < -0.88)
Using z table,
= 0.8106 - 0.1894
= 0.6212
Probability = 0.6212
b ) n = 50
= = 273
= / n = 100/ 50 = 14.1421
P (x > 289 )
= 1 - P (x < 289 )
= 1 - P ( - ) / ) < ( 289 - 273 / 14.1421)
= 1 - P ( z < 16 / 14.1421 )
= 1 - P ( z < 1.13 )
Using z table
= 1 - 0.8708
= 0.1292
Probability = 0.1292
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