Business Stats Take home exam Ch 5-8 7) Wage ve a mean of $22.91 and standard de
ID: 3309196 • Letter: B
Question
Business Stats Take home exam Ch 5-8 7) Wage ve a mean of $22.91 and standard deviation of $2.83 per hour. Assume wages are normally to earn: a) More than 25% (approximately) of all employees? b) The 99h percentile (approximately) of all employees? c) f picked distributed (pretty close). What must your hourly wage be if you want 77andom, what is the probablity that the employee will earmn between $17.25 at random, what is the probability that the employee will cam between $20.08 8) The average annual published cost of tuition, room and board at private, 4 year university and $28.57 and $31.40 for the academic year 2017-2018 is $46,950 according to the "Trends in College Phcing 2017 report of the College Board. A group of financial aid administrators believe that the average cost is lower. A study conducted using a sampling of 25 private universities showed that the average cost per year is $42,874 with a standard deviation of $1,818. a) How are the means distributed? Explain. b) Find the 95% confidence interval for the average college cost according to administrators. c) How would this be distributed if the sample size ere increased to 225? d) Find the confidence interval with a sample size of 225. e) How many would you need to sample is you have an allowable error of $690? the 4th quarter of 2016. We sampled 2,00 homeowners and found 8 vacant homes. Based on the information above: a) Find a 90% confidence interval for homeownership vacancies. b) Based on the data, is this a representative sample if the census data is correct? 9) According to the Census Bureau, the homeownership vacancy rate in the US is 1.8%in 80 SExplanation / Answer
mean = 22.91
std.dev. = 2.83
a)
z - value = 0.67448975
xbar = mean + z*sigma
xbar = 22.91 + 0.67448975*2.83
xbar = 24.81880599
b)
z-value = 2.33
xbar = mean + z*sigma
xbar = 22.91 + 2.33*2.83
xbar = 29.5039
c)
P(17.25 < X < 28.57)
= P((17.25 - 22.91)/2.83 < z < (28.57 - 22.91)/2.83)
= P(-2 < z < 2)
= 0.9545
d)
P(20.08 < X < 31.4)
= P((20.08 - 22.91)/2.83 < z < (31.4 - 22.91)/2.83)
= P(-1 < z < 3)
= 0.83999
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