Download ENGR 323 Heidi M. Gehlhaar Spring 1998 May 1, 1998 HW #14 1/2

ENGR 323
Spring 1998
HW #14
Heidi M. Gehlhaar
May 1, 1998
1/2
Problem 5-97
5-97) Problem Statement:
The weight of a small candy is normally distributed with a mean of 0.1 ounce and
a standard deviation of 0.01 ounce. Suppose that 16 candies are placed in a
package and that the weights are independent.
a.) What are the mean and variance of package net weight?
b.) What is the probability that the net weight of a package is less than 1.6
ounces?
c.) If 17 candies are placed in each package, what is the probability that the net
weight of a package is less than 1.6 ounces?
Problem Solution:
We will begin with variable definition:
Let Wi = Weight of the ith small candy in ounces
Wi ~ N(µW = 0.1, σW = 0.01)
Let X = Weight of a package in ounces containing 16 candies
Because X is just 16 of W, X is just a linear combination of W and is normally
distributed and has the same type of random variables with different parameters:
X = W1 + W2 + W3 + ... + W16
Wi are iid
X ~ N(µX = ?, σX = ?)
a.) Here we will find the parameters of the distribution of X, including the mean and
variance. To do this, we will use the general rule we learned in class about linear
combinations of random variables (page 273):
For A = c1B1 + c2B2 +...+ cnBn,
E(A) = c1E(B1) + c2E(B2) +...+ cnE(Bn)
V(A) = c12(B1) + c22(B2) + ...+ cn2(Bn)
For our problem,
E(X) = E(W1) + E(W2) + ... + E(W16)
= 16(0.1)
E(X) = 1.6 ounces
V(X) = V(W1) + V(W2) + ... + V(W16)
= 16(0.012)
V(X) = 0.0016 ounce
where E(W) = µW = 0.1
where V(W) = σW2 = 0.012
ENGR 323
Spring 1998
HW #14
Heidi M. Gehlhaar
May 1, 1998
2/2
Problem 5-97 Continued
It is very important to note that we use the variance of the random variables,
not the standard deviation! Many mistakes can be made by accidentally using
the standard deviation instead of the variance in a linear transformation. Beth
pointed out that it is just as important to remember as a2+b2=c2 and NOT a+b=c
for a right triangle! The two concepts are very similar and equally messed up by
unknowing students. The expected value can be changed linearly but the standard
deviation cannot. When performing problems such as this one, be sure to note if
your standard deviation changed by a reasonable amount. When doing a linear
combination of random variables, sigma will increase, but not as much as it would
if increased linearly with the expected value. An extreme change in expected
value will definitely give you a change in standard deviation but not as
dramatically as might be expected.
b.) The probability that a package weighs less than 1.6 ounces can be written as
P(X<1.6). Since we have already shown that X is distributed normally, we can
standardize and use our Z table to solve this portion of the problem.
1.6 − 1.6 
P(X<1.6) =  Z <


0.04 
= P(Z<0)
P(X< 1.6) = 0.5
50% makes sense seeing as the problem was really asking for the probability that
a package weighs less than the mean. We would expect to find packages that
weigh less than the mean half of the time and packages that weight more than the
mean the other half of the time. (Remember that, with continuous random
variables, P(X=x) = 0.)
c.) This portion of the problem is a combination of parts a and b but with packages
containing 17 candies instead of 16 candies. We need a new random variable:
Let Y = Weight of a package in ounces containing 17 candies
Y = W1 + W2 + W3 + ... + W17
Wi are iid
As before,
E(Y) = E(W1) + E(W2) + ... + E(W17)
where E(W) = µW = 0.1
= 17(0.1)
E(Y) = 1.7 ounces
And
V(Y) = V(W1) + V(W2) + ... + V(W16)
where V(W) = σW2 = 0.012
= 17(0.012)
V(Y) = 0.0017 ounce
As explained earlier, the distribution of Y will be normal:
Y ~ N(µY = 1.7, σY = 0.0412)
Now we find the desired probability by standardizing:
1.6 − 1.7 

P(Y<1.6) =  Z <


0.0017 
= P(Z < -2.43)
Then use the Z table:
P(Y<1.6) = 0.008