How can I solve this problem for normal distribution with two populations?

Question

Suppose the net weight of a package (ie 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 758g and standard deviation 56.5g, while the weight of the packaging of the package (say Y) has a normal distribution with mean 49g and standard deviation 3.3g.
(a) 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 857.37g.

Answer 1

Hi,



(a) By adding the mean weights and standard deviations, you'd get a total mean weight of 807 g and a total standard deviation of 59.8 g.



(b) To find the probability that the total weight of the package exceeds 857.37g, use the normalcdf( command on a T-I83 found under [2nd][VARS] under DISTR. For this command enter the following data:



normalcdf(lower bound, upper bound, mean, standard deviation)



In this case, enter:

normalcdf(857.37,9999, 807, 59.8)



This solves to:

normalcdf(857.37,9999, 807, 59.8) = .1998



The probability that the total weight of the package exceeds 857.37g is 19.98%.



I hope that helps!! :-)

Answer 2

Check out

In that article you will see the following property of the normal distribution:



If X is distributed with normal probability density function N(mu,v^2),

and Y is distributed with normal probability density function N(nu,u^2),

and if X and Y are independent random variables then

Z = X+Y is distributed normally with probability density function N(mu+nu,v^2+u^2),

where mu and nu are the means values for the normal distributions and v^2 and u^2 are the variances.

Thus one must add the means together and add the variances together, *not* add the standard deviations.



In our case Z must then be distributed normally with mean:

mean = 758 + 49 = 807 grams

and with variance:

var = 56.5^2 + 3.3^2 = 3192.25 + 10.89 = 3203.14

and from the variance we can get the standard deviation:

stdev = sqrt(3203.14) ~ 56.5963



The probability that the total weight Z exceeds 857.37 grams is then found by first calculating the standard normal lookup factor z:

z = (Z - mean)/std = (857.37 - 807)/56.5963 = 0.8899875

Then one can use z to look up the cumulative probability value from a table of values for the cumulative standard normal distribution.

That is,

NormCDF(x > 857.37)

= StdNormCDF(z > 0.8899875)

= StdNormCDF(z < -0.8899875)

= 0.1867

where I have looked up the value from the table on the web site .



Thus the probability that the toal weight of the package exceeds 857.37 grams is 0.1867

Answer 3

Given: X ~ N(758, 56.5^2)

Given: Y ~ N(49, 3.3^2)



(a) Let T = total weight of the package

Then T = X + Y

Since X, Y are independent (<-- this condition is extremely important) normal random variables,

E(T) = E(X) + E(Y) = 758 + 49 = 807 and

Var(T) = 56.5^2 + 3.3^2 = 3203.14, so that

s.d. of T = sqrt(3203.14) = 56.6



(b) P(T > 857.37) = 0.187, with a graphic calculator.