Probability Homework Help

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The following table gives the probability distribution of X, the number of defective products in a production line in a day.
x P(X=x)

0 0.04

1 0.19

2 0.35

3 0.26

4 0.16

a. Evaluate P(|X - expected value| >= standard dev)
b. The cost of maintaining the machine at the production line is given by C = 50 + 3x + 2x^2. Find the mean maintaining cost.
 
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  • #2
Ray Vickson
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The following table gives the probability distribution of X, the number of defective products in a production line in a day.
x P(X=x)

0 0.04

1 0.19

2 0.35

3 0.26

4 0.16

a. Evaluate P(|X - expected value| >= standard dev)
b. The cost of maintaining the machine at the production line is given by C = 50 + 3x + 2x^2. Find the mean maintaining cost.

Show your work. What have you done so far? Where are you stuck?

RGV
 
  • #3
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The following table gives the probability distribution of X, the number of defective products in a production line in a day.
x P(X=x)

0 0.04

1 0.19

2 0.35

3 0.26

4 0.16

a. Evaluate P(|X - expected value| >= standard dev)
b. The cost of maintaining the machine at the production line is given by C = 50 + 3x + 2x^2. Find the mean maintaining cost.

I only managed this much:

Expected value = 2.31
Standard deviation = 1.074

P( |X - 2.31| >= 1.074)
=P( -1.074 >= (X - 2.31) >= 1.074 )
=P( 1.236 >= X >= 3.384)

I'm stuck here. For question b, I assume I have to calculate the expected value of C by substituting the variables x and x^2 with their respective expected values?
 
  • #4
vela
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I only managed this much:

Expected value = 2.31
Standard deviation = 1.074

P( |X - 2.31| >= 1.074)
=P( -1.074 >= (X - 2.31) >= 1.074 )
=P( 1.236 >= X >= 3.384)
This doesn't make sense since 1.236 clearly isn't less than or equal to 3.384.

The condition |X - 2.31| ≥ 1.074 is equivalent to (X-2.31 ≥ 1.074) or (X-2.31 ≤ -1.074), so
[tex]P(\lvert X-2.31 \rvert \ge 1.074) = P(X-2.31 \ge 1.074) + P(X-2.31 \le -1.074)[/tex]
For question b, I assume I have to calculate the expected value of C by substituting the variables x and x^2 with their respective expected values?
Yes, because E(C) = E(50 + 3x + 2x^2) = E(50) + E(3x) + E(2x2) = 50 + 3 E(x) +2 E(x2)
 
  • #5
Ray Vickson
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I only managed this much:

Expected value = 2.31
Standard deviation = 1.074

P( |X - 2.31| >= 1.074)
=P( -1.074 >= (X - 2.31) >= 1.074 )
=P( 1.236 >= X >= 3.384)

I'm stuck here. For question b, I assume I have to calculate the expected value of C by substituting the variables x and x^2 with their respective expected values?

|X - 2.31| >= 1.074 <--> either X - 2.31 >= 1.074 or 2.31 - X >= 1.074.

RGV
 
  • #6
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So is this how it goes, based on the given values of x and P(X=x):

P(|X-2.31| >= 1.074)
= P( X-2.31 >= 1.074) + P( 2.31-X >= 1.074)
= P( X >= 3.384) + P( X <= 1.236)
= 0.16 + 0.19 + 0.04
= 0.39
 
  • #7
vela
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Yes, that's right.
 

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