Means and Variances for Linear Transformations; more on Normal Probabilities Sup
ID: 3152139 • Letter: M
Question
Means and Variances for Linear Transformations; more on Normal Probabilities Suppose the net weight of a package (i.e. the weight of its contents) and the weight of the packaging of the package can be considered to be independent random variables. Assume the net weight of the package (say X) has a normal distribution with mean 766g and standard deviation 58.2g, while the weight of the packaging of the package (say Y) has a normal distribution with mean 24g and standard deviation 3.8g. Write down an expression for the total weight of the package in terms of X and Y. Find the mean and standard deviation of the total weight of the package. (b) Find the probability that the total weight of the package exceeds 871.65g. More on the Normal Quantiles In considering the reliability of domestic-waste biofilm reactors, suppose the substrate concentration (mg/cm^3) of influent to a reactor is normally distributed with mean 0.63. Suppose circumstances can be changed so that the variability of the concentration can be set; that is, the standard deviation a can be set at a pre-determined value. Find the value of a so that there is a probability of 0.9 that the concentration exceeds 0.5634.Explanation / Answer
Let X = net weight of the package.
X ~ Normal (mean = 766, sd = 58.2)
Y = weight of the packeging of the package.
Y ~ Normal (mean = 24, sd = 3.8)
Total weight (W) = X+Y ~ Normal(mean = 766 + 24 = 790, sd = 58.2 + 3.8 = 62)
W ~ Normal (mean = 790, sd = 62)
Here we have to find P(W > 871.65).
First convert w = 871.65 into z-score.
z = (w - mean) / sd
z = (871.65 - 790) / 62 = 1.32
That is now we have to find P(Z > 1.32).
This probability we can find by using EXCEL.
syntax is,
= 1 - NORMSDIST(z)
where z is test statistic value.
P(Z > 1.32) = 0.0939
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Given that,
mean = 0.63
sd = ?
P(X > 0.5634) = 0.9
now convert x = 0.5634 into z-score.
z = (0.5634 - mean) /sd = (0.5634 - 0.63) / sd
P(Z > (0.5634 - 0.63) / sd ) = 0.9
That is now we have to find inverse value of normal distribution when probability is given.
This also we can done using EXCEL.
syntax is,
= NORMSINV(probability)
where probability = 1 - 0.9 = 0.1
This will give me value -1.28.
Now compare it with (0.5634 - 0.63) / sd.
That is (0.5634 - 0.63) / sd = -1.28
-0.0666 / sd = -1.28
-0.0666 / -1.28 = sd
sd = 0.052