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STAT 350 – Fall 2008
Homework #5 – Solution
through Lecture #9
Text Exercises
Chapter 7, Problem #30
7.30
()
22
4.89
6.43
26.99 35.76
1.96
68
74
−±
+
= (10.6401, 6.8999)
Note:
(6.8999, 10.6401) is also acceptable, this is just the 95% CI for
21
μ
−
instead of
12
−
Then answer the following problems
1. According to www.mms.com
, 20% of milk chocolate M&M's are red.
Sally will eat one
snack pack of milk chocolate M&M's, for a total of 23 M&M's.
Let
p
1
denote the proportion
of M&M's Sally will eat that are red.
a.
What is the expected value of
p
1
?
E(
p
1
) =
π
= 0.20
b.
What is the standard deviation of
p
1
?
2
1
0.2 1 0.2
0.0834058
23
SD p
n
ππ
−−
==
=
c.
Find the exact probability that
p
1
< 0.15.
Hint:
you should not be using the normal
distribution for this problem!
We cannot assume that p
1
is approximately normally distributed, because n
π
= 4.6 < 5
.
What we can do instead is to convert the question to a question about the number
of red
M&M's in the bag, and answer the question using the Binomial distribution.
For
p
1
to equal 0.15, there would have to be 0.15(23) = 3.45 red M&M's in the bag.
Any
less than that and
p
1
will be less than 0.15.
Let
X
= the number of red M&M's in this bag
X
~ Binomial(n = 23,
π
= 0.20)
We need to find P(
X
< 3.45) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
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This note was uploaded on 02/23/2009 for the course STAT 350 taught by Professor Staff during the Fall '08 term at Purdue University.
 Fall '08
 Staff

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