Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN: 9781133382119
Author: Swokowski
Publisher: Cengage
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![A shipping company handles containers in three different sizes: (1) 27 ft³ (3 × 3 × 3), (2) 125 ft³, and (3) 512 ft³. Let X; (i = 1, 2, 3) denote the number of type i containers shipped during a given
week. With µ; = E(X;) and σ, 2² = V(X;), suppose that the mean values and standard deviations are as follows:
= 230
M1
°₁ = 11
M2 = 260 Из = 150
02
=
13
7
03
(a) Assuming that X1, X2, X3 are independent, calculate the expected value and variance of the total volume shipped. [Hint: Volume
expected value
variance
ft³
ft6
=
27X1 + 125X2 + 512X3.]
(b) Would your calculations necessarily be correct if the X,'s were not independent? Explain.
○ The expected value would be correct, but the variance would not be correct.
Neither the expected value nor the variance would be correct.
Both the expected value and the variance would be correct.
The expected value would not be correct, but the variance would be correct.](https://content.bartleby.com/qna-images/question/b898cad9-5347-4e0a-a74d-32f84bfad0f6/fe8af8a6-43b6-4497-873d-25986d5aaa11/s6gbo0e_thumbnail.png)
Transcribed Image Text:A shipping company handles containers in three different sizes: (1) 27 ft³ (3 × 3 × 3), (2) 125 ft³, and (3) 512 ft³. Let X; (i = 1, 2, 3) denote the number of type i containers shipped during a given
week. With µ; = E(X;) and σ, 2² = V(X;), suppose that the mean values and standard deviations are as follows:
= 230
M1
°₁ = 11
M2 = 260 Из = 150
02
=
13
7
03
(a) Assuming that X1, X2, X3 are independent, calculate the expected value and variance of the total volume shipped. [Hint: Volume
expected value
variance
ft³
ft6
=
27X1 + 125X2 + 512X3.]
(b) Would your calculations necessarily be correct if the X,'s were not independent? Explain.
○ The expected value would be correct, but the variance would not be correct.
Neither the expected value nor the variance would be correct.
Both the expected value and the variance would be correct.
The expected value would not be correct, but the variance would be correct.
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