# Probability Homework Help

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.

Last edited:

Ray Vickson
Homework Helper
Dearly Missed
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

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?

vela
Staff Emeritus
Homework Helper
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
$$P(\lvert X-2.31 \rvert \ge 1.074) = P(X-2.31 \ge 1.074) + P(X-2.31 \le -1.074)$$
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)

Ray Vickson
Homework Helper
Dearly Missed
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

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

vela
Staff Emeritus