attempt Q5 FAConditional Format as NeutralFormatting TableA-OMerge & Center6 %.0000 .0CalculationCheck CeAlignmentNumberStyleshether silicon chip sales are independent of where the XYZ country's economy is in the business cycle, data have been collected on the weekly salesJKMQR.Upercent, and the sample standard deviation was1.39 percent. Is it reasonable toconclude (at a=0.01) that wealth-market interest rates declined during 2015?A tire manufacturer warranties its tires to last at leastQ520275 miles or "you get a new set of tires."In its experience, a set of these tires lasts on average26275 miles with standard deviation5110 miles.Assume that the wear is normally distributed. The manufacturer profits Rs. 200 on each set sold, andreplacing a set costs the manufacturer Rs. 400.a) What is the probability that a set of tires wears out before20275 miles?20275 miles?b) What is the probability that a set of tires wears out beforec) What is the cutoff value in terms of miles of the top24275 miles and after55% wearing out tires.Q6AC, Ltd., a supplier of operating system software for personal computers, was planning the initial publicoffering of its stock in order to raise sufficient working capital to finance the development of a radically new,seventh-generation integrated system. With current earnings of $1.61 a share, AC and its underwriters werecontemplating an offering price of $21, or about 13 times earnings. In order to check the appropriateness of thisprice, they randomly chose seven publicly traded software firms and found that their average price/earnings ratio1.85. At a 0.02, can AC conclude that the stocks15.75 ?was14.4 , and the sample standard deviation wasof publicly traded software firms have an average P/E ratio that is significantly different fromWhat are the confidence limits of acceptance region?

Name: attempt Q5 FAConditional Format as NeutralFormatting TableA-OMerge & C
Uploaded: 2024-03-20T06:38:44.000Z
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Description: attempt Q5 FAConditional Format as NeutralFormatting TableA-OMerge & Center6 %.0000 .0CalculationCheck CeAlignmentNumberStyleshether silicon chip sales are independent of where the XYZ country's economy is in the business cycle, data have been collec

Question

attempt Q5 FAConditional Format as NeutralFormatting TableA-OMerge & Center6 %.0000 .0CalculationCheck CeAlignmentNumberStyleshether silicon chip sales are independent of where the XYZ country's economy is in the business cycle, data have been collected on the weekly salesJKMQR.Upercent, and the sample standard deviation was1.39 percent. Is it reasonable toconclude (at a=0.01) that wealth-market interest rates declined during 2015?A tire manufacturer warranties its tires to last at leastQ520275 miles or &#34;you get a new set of tires.&#34;In its experience, a set of these tires lasts on average26275 miles with standard deviation5110 miles.Assume that the wear is normally distributed. The manufacturer profits Rs. 200 on each set sold, andreplacing a set costs the manufacturer Rs. 400.a) What is the probability that a set of tires wears out before20275 miles?20275 miles?b) What is the probability that a set of tires wears out beforec) What is the cutoff value in terms of miles of the top24275 miles and after55% wearing out tires.Q6AC, Ltd., a supplier of operating system software for personal computers, was planning the initial publicoffering of its stock in order to raise sufficient working capital to finance the development of a radically new,seventh-generation integrated system. With current earnings of $1.61 a share, AC and its underwriters werecontemplating an offering price of $21, or about 13 times earnings. In order to check the appropriateness of thisprice, they randomly chose seven publicly traded software firms and found that their average price/earnings ratio1.85. At a 0.02, can AC conclude that the stocks15.75 ?was14.4 , and the sample standard deviation wasof publicly traded software firms have an average P/E ratio that is significantly different fromWhat are the confidence limits of acceptance region?

Accepted Answer

Given
μ=26275σ=5110
 
Use normalization as,
z=x-μσ