A certain store sells only T-shirts. Let the random variable X represent the number of T-shirts bought from the store on any day, with a mean of 75 and a standard deviation of 20. Let the random variable Y be the total revenue from this store on a randomly selected day. If the store charges $15 per T-shirt, what are the mean and standard deviation of Y

Answers

Answer 1

Answer:

[tex](a)\ \mu_y = 1125[/tex]

[tex](b)\ \sigma_y = 300[/tex]

Step-by-step explanation:

Given

[tex]\mu_x = 75[/tex] -- Mean of T-shirts

[tex]\sigma_x = 20[/tex] -- Standard deviation of T-shirts

[tex]Rate = \$ 15[/tex]

Solving (a): The mean of the revenue [tex](\mu_y)[/tex]

To solve this, we use:

[tex]\mu_y = Rate * \mu_x\\[/tex]

This gives:

[tex]\mu_y = 15 * 75[/tex]

[tex]\mu_y = 1125[/tex]

Solving (b): The standard deviation of the revenue [tex](\sigma_y)[/tex]

To solve this, we use:

[tex]\sigma_y = \sqrt{Rate^2 * \sigma_2^2}[/tex]

This gives:

[tex]\sigma_y = \sqrt{15^2 * 20^2}[/tex]

[tex]\sigma_y = \sqrt{90000}[/tex]

[tex]\sigma_y = 300[/tex]


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Answer:

table C

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There is not sufficient evidence to warrant the rejection of the claim that the mean weight of cereal is atleast 14 oz

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Answers

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Answer:

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Answer:

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Answers

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Answers

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Answers

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Answers

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