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A random sampling of a company's monthly operating expenses for n=36 months produced a sample mean of $5474 and a stan-dard deviation of S 764. Find a 90% upper confidence bound for the company's mean monthly expenses.

User RaptoX
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Answer with explanation:

Mean of the sample(m) = $ 5474

Standard deviation of the sample (S)=764

Number of observation(n)=36


Z_{90 \text{Percent}}=Z_(0.09)=0.5359


z_(score)=(\Bar x-\mu)/((S)/(√(n)))\\\\0.5359=(5474- \mu)/((764)/(√(36)))\\\\0.5359=(5474- \mu)/((764)/(6))\\\\764 * 0.5359=6 * (5474- \mu)\\\\409.4276=32844-6 \mu\\\\6 \mu=32844 -409.4276\\\\ 6 \mu=32434.5724\\\\ \mu=(32434.5724)/(6)\\\\ \mu=5405.76

So, Mean Monthly Expenses of Population =$ 5405.76, which is 90% upper confidence bound for the company's mean monthly expenses.

User Mayank Chandak
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