The proportion of time per day that all checkout counters in a supermarket are busy is a random variable Y with a density function given by |cy²(1 – y)5, 0 µ + 20). (Round your answer to four decimal...

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The proportion of time per day that all checkout counters in a supermarket are busy is a random variable Y with a density function<br>given by<br>|cy²(1 – y)5, 0 <ys 1,<br>f(y) :<br>elsewhere.<br>(a) Find the value of c that makes f(y) a probability density function.<br>C = 0.01190<br>(b) Find E(Y). (Use what you have learned about the beta-type distribution.)<br>E(Y)<br>.1666<br>(c) Calculate the standard deviation of Y. (Round your answer to four decimal places.)<br>=<br>(d) Find P(Y > µ + 20). (Round your answer to four decimal places.)<br>

Extracted text: The proportion of time per day that all checkout counters in a supermarket are busy is a random variable Y with a density function given by |cy²(1 – y)5, 0 µ + 20). (Round your answer to four decimal places.)

Jun 07, 2022

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The proportion of time per day that all checkout counters in a supermarket are busy is a random variable Y with a density function given by |cy²(1 – y)5, 0 µ + 20). (Round your answer to four decimal...

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