Mathematics, 16.04.2020 00:28 tsbrunson13

A particular variety of watermelon weighs on average 21.3 pounds with a standard deviation of 1.07 pounds. Consider the sample mean weight of 90 watermelons of this variety. Assume the individual watermelon weights are independent. a. What is the expected value of the sample mean weight

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Answer from: jeffro198004

a) expected value = 21.3 pounds

b) standard deiation of the sample mean weight = 0.1128

c) P(x>21.3)=0.5000

d) C=21.4665

Step-by-step explanation:

The question is incomplete:

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places. 2

c. What is the approximate probability the sample mean weight will be greater than 21.3? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question. 3

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.93? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question. 4

We have the parameters of the population of melon weights (mean: 21.3 pounds, standard deviation: 1.07 pounds) and we have to describe the sampling distribution of sample means, when the sample size is n=90.

The mean of the sampling distribution (the expected value of the sample mean weight) is equal to the population mean: 21.3 pounds.

The standard deviation of the sampling distribution is calculated as:

$\sigma_M=\sigma/\sqrt{n}=1.07/\sqrt{90}=1.07/9.49=0.1128$

As the mean of the sampling distribution is 21.3, the probability that the sample mean exceeds 21.3 pounds is P=0.5000.

Now, to calculate C so that P(x<c)=0.93 we have to look up for the z-scor z* so that P(z<z*)=0.93.

According to the standard normal distribution, this z-score is z*=1.4758.

Then, we can convert this value to our specific distribution and calculate c as:

$C=\mu_M+z\cdot \sigma_M=21.3+1.4758*0.1128=21.3+0.1665=21.4665$

Answer from: Quest

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

Answer from: Quest

step-by-step explanation:

Which of the following statements are true for the image of a triangles after a dilation that has a

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