Question

Biological Oxygen Demand (BOD) is an index of pollution that is monitored in the treated effluent (the liquid waste produced during manufacturing, typically discharged into nearby rivers) of paper mills (as well as other manufacturing buildings) on a regular basis. Suppose that the BOD (in pounds per day) is recorded at a particular paper mill during the spring and summer months. If a BOD of 3526 was 1.3 standard deviations above the mean and a BOD of 3001 was -2.6 standard deviations below the mean, what is the mean BOD of the paper mill, to the nearest pound?

Answer 1

let mean is a and standard deviation is b

therefore 1.3 standard deviations above the mean value =a+1.3b =3526 ........(1)

and 2.6 standard deviations below the mean value =a-2.6b =3001 ........(2)

equation(1) -equation(2)

3.9b =3526-3001

b=134.6154

therefore a = mean BOD of the paper mill =3526-1.3*134.6154 =3351.00

Answer 2

To find the mean BOD of the paper mill, we need to first determine the standard deviation (σ) from the given information and then use it to calculate the mean.

Let's denote the mean BOD as μ (mu).

Given information: BOD of 3526 was 1.3 standard deviations above the mean: BOD1 = μ + 1.3σ BOD of 3001 was -2.6 standard deviations below the mean: BOD2 = μ - 2.6σ

Now, we can set up two equations using the given information:

1. 3526 = μ + 1.3σ

2. 3001 = μ - 2.6σ

Next, let's solve these equations to find the values of μ and σ.

Subtract equation (2) from equation (1):

3526 - 3001 = μ + 1.3σ - (μ - 2.6σ) 525 = 3.9σ

Now, divide both sides by 3.9 to solve for σ:

σ = 525 / 3.9 σ ≈ 134.62

Now that we have the value of σ, we can use either equation (1) or (2) to find the mean μ. Let's use equation (1):

3526 = μ + 1.3σ

Substitute the value of σ:

3526 = μ + 1.3(134.62)

Now, solve for μ:

μ = 3526 - 174.806 μ ≈ 3351.194

So, the mean BOD of the paper mill is approximately 3351 pounds per day (rounded to the nearest pound).