(1) According to the Australian Department of Industry, Tourism and Resources (DITR), 8.6% of the total employment in NSW is related to manufactured exports. A sample of 200 employees in NSW is...

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Answered Same DayDec 23, 2021

Answer To: (1) According to the Australian Department of Industry, Tourism and Resources (DITR), 8.6% of the...

Robert answered on Dec 23 2021
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(1) According to the Australian Department of Industry, Tourism and Resources (DITR), 8.6% of the total employment in NSW is related to manufactured exports. A sample of 200 employees in NSW is randomly selected. If X is the number of employees in the sample with jobs related to manufactured exports, then the standard deviation of X is _______________.
Answer: 3.96
We have n = 200 and p = 0.086
Standard deviation = √ n p (1 – p)
= √200*0.086*(1 – 0.086)
= √15.7208
= 3.96
(2) Suppose you are working with a data set that is normally distributed with a mean of 400 and a standard deviation of 20. Determine the value of x such that only 1% of the values are greater than x. (how to find the value of x with z score)
Answer: 446.6
We have µ = 400 and σ = 20
Using z-tables, the z-score with p > 0.01 is 2.33
Z = (X - µ)/σ
2.33 = (X – 400)/20
2.33*20 + 400 = X
X = 446.6
(3) Penny Bauer, Chief Financial Officer for Harrison Haulage, suspects irregularities in the payroll system. If 10% of the 5000 payroll vouchers issued since 1 January 2005, have irregularities, the probability that Penny’s random sample of 200 vouchers will have a sample proportion .06 and .14 is ___________.
Answer: 0.9412
We have n = 200 and p = 0.10
The respective Z-score with p’ = 0.06 is
Z = (p’ – p)/√ p (1 – p)/n
= (0.06 – 0.10)/ √0.10*0.90/200
= -1.89
The respective Z-score with p’ = 0.14 is
Z = (p’ – p)/√ p (1 – p)/n
= (0.14 – 0.10)/ √0.10*0.90/200
= 1.89
Using Z-tables, the probability is
P [-1.89 < Z < 1.89] = 0.9706 – 0.0294
= 0.9412
(4) Catherine Cho, Director of Marketing Research, needs a sample of Darwin households to participate in the testing of a new toothpaste. If 40% of the households in Darwin prefer the new toothpaste, the probability that Catherine’s random sample of 300 households will have a sample proportion between 0.35 and 0.45 is ___________.
Answer: 0.9232
We have n = 300 and p = 0.40
The respective Z-score with p’ = 0.35 is
Z = (p’ – p)/√ p (1 –...
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