Download MA 1125 Lecture 12 - Mean and Standard Deviation for the Binomial

MA 1125 Lecture 12 - Mean and Standard Deviation for the Binomial Distribution
Monday, February 24, 2014.
Objectives: Mean and standard deviation for the binomial distribution.
1. Mean and Standard Deviation of the Binomial Distribution
One of our binomial distributions involved tossing a coin three times. According to our
ideas about probability, we expect half of our tosses to come up heads. Therefore, we would
expect, on average, that we would get 1.5 heads in three tosses. The mean for this particular
binomial distribution should be µ = 1.5 heads. What should the standard deviation be?
What we’ll do today is to figure out a way of computing the mean for all binomial distributions. We won’t go into as much detail, but the standard deviation can be calculated using
the same ideas.
2. Computing the mean and standard deviation
We intuitively know what the mean should be for the binomial distribution just discussed.
But how could we compute it from a formula? The formula we have only works for finite
data sets, and our population here is infinite. Here’s one way.
Recall the probabilities.
(1)
Number of heads
Probability
0
1
8
1
3
8
2
3
8
3
1
8
The probabilities for tossing a coin three times are all fractions with 8 in the denominator.
Let’s suppose we repeated the experiment eight times. In theory, since P (0 heads) = 81 , we
would expect to see 0 heads once. Since P (1 head) =
3
,
8
we would expect to see 1 head
three times. Similarly, we would expect 2 heads three times, and 3 heads only once.
1
2
Therefore, if we repeated the experiment eight times, our theoretical expectations would be
as follows.
x
0
1
1
(2)
1
2
2
2
3
Our variable x takes on the values 0 once, 1 and 2 three times, and 3 once. With only eight
numbers, we know how to compute the mean. We add them up and divide by n,
P
n
x
.
x
0
1
1
1
(3)
2
2
2
3
12
µ=
12
8
= 1.5
This is a theoretical computation that applies to all conceivable repetitions of the experiment, so we’re doing a computation on a population. That means that we’re computing the
parameter µ (the population mean). This is really a short cut to listing out one billion 0’s,
three billion 1’s, three billion 2’s, and one billion 3’s. We’ll use N = 8, but this could be
N = 8 billion. Actually, it’s a lot bigger than billions.
Again, since we’re only dealing with eight numbers, we can compute the variance and
standard deviation using the formulas we have for populations.
(4)
2
σ =
P
(x − µ)2
.
N
MA 1125 Lecture 12 - Mean and Standard Deviation for the Binomial Distribution
3
For the population variance, we divide by N (since 8 billion or more minus one doesn’t
really matter).
(5)
x
x−µ
(x − µ)2
0
−1.5
2.25
1
−0.5
0.25
1
−0.5
0.25
1
−0.5
0.25
2
0.5
0.25
2
0.5
0.25
2
0.5
0.25
3
1.5
2.25
12
µ=
12
8
6.00
2
= 1.5
6.00
8
σ =
= 0.75
√
σ = 0.75 = 0.87
The variance and standard deviation tell us how spread out the possible outcomes are.
3. A formula for the mean
Among mathematicians, the way we computed the mean, variance, and standard deviation
in the last section wouldn’t be considered terribly cool. Let’s look at another way.
Consider the mean, µ. Essentially, we did the computation
(6)
µ=
0+1+1+1+2+2+2+3
.
8
Instead of adding the 1’s and 2’s three times, we could have multiplied by 3. This would
have been
(7)
µ=
0·1+1·3+2·3+3·1
.
8
Also, instead of dividing by 8 at the end, we could just divide each term by 8 to get
(8)
µ=0·
1
3
3
1
+1· +2· +3· .
8
8
8
8
In other words, to find the mean, we can take each outcome, multiply by its probability,
and then add them all up. This gives us a formula for finding the mean of a probability
distribution.
(9)
µ=
X
(x · P (x)).
This formula works for all probability distributions (well, at least the ones with a finite
number of outcomes). In the specific case of a binomial distribution, the formula can be
4
simplified quite a bit. If p is the probability for the thing you’re counting, and n is the
number of trials, then for a binomial distribution
(10)
µ = n · p.
For tossing a coin three times, which is what we started today’s lecture with, n = 3 and
p = 12 . The mean, therefore, would be
(11)
µ=3·
1
= 1.5,
2
which is just what we got.
We won’t, but we could go through the computations for finding the variance and standard
deviation for a binomial distribution in general. The formula for the variance for a binomial
distribution would simplify to the following.
σ 2 = n · p · q,
(12)
where q is the probability of a failure. In the case of rolling a die, q = 1 − p is the probability
of not getting a 1. We usually describe p as the probability of a success, and q = 1 − p as the
probability of a failure. The formulas for the variance and standard deviation of a binomial
distribution are
σ 2 = npq
(13)
σ=
√
npq
4. Quiz 12
For the binomial experiment, where we roll a die ten times, and count the 1’s and 3’s, find
the following.
1.
n.
2.
p. (Write as a fraction.)
3.
q. (Write as a fraction.)
4.
µ. (Round to two decimal places.)
5.
σ. (Round to two decimal places.)
Suppose I take a twenty-question multiple choice test, where each question has four answers
to choose from. A success is guessing a correct answer. Find the following.
6.
n.
MA 1125 Lecture 12 - Mean and Standard Deviation for the Binomial Distribution
7.
p. (Write as a fraction.)
8.
q. (Write as a fraction.)
9.
µ. (Round to two decimal places, if necessary.)
10.
σ. (Round to two decimal places, if necessary.)
5
5. Homework 12
Suppose you take a thirty-question multiple choice test by guessing, and each question has
five choices. Find the following.
1.
n.
2.
p. (Write as a fraction.)
3.
q. (Write as a fraction.)
4.
µ. Round to two decimal places.
5.
σ. Round to two decimal places.
6.
What score is three standard deviations above the mean? (That is, what score corre-
sponds to a z-score of z = 3.00?) Round to two decimal places.
7.
If three standard deviations above the mean is the best you could reasonably hope for
(but you still have to be pretty lucky to do that well), do you have much hope of getting 15
correct out of 30?
Remember that my brother is a 60% freethrow shooter. If he tries 15 freethrows, find the
following. Making a freethrow is a success, of course.
8.
Find µ. (Round to two decimal places.)
9.
Find σ. (Round to two decimal places.)
10.
If three standard deviations below the mean is the worst we’d normally see him shoot,
how many made freethrows is that? (Round to the nearest whole number.)
Answers on next page.
6
Quiz Answers:
1) n = 10.
2) p = P (1 OR 3) = 13 .
3) q = 23 .
4) µ = np = 10 · 13 = 3.33.
q
q
√
= 1.49.
5) σ = npq = 10 · 31 · 23 = 20
9
HW Answers:
1) n = 30.
2) p = 15 .
3) q = 45 .
4) µ = np = 30 · 15 = 6.
q
√
5) σ = npq = 30 · 51 ·
4
5
= 2.19
6) µ + 3 · σ = 6 + 3 · 2.19 = 12.57 correct.
7) Outside of 3 standard deviations is extremely unusual, so no.
8) µ = np = 15 · 0.60 = 9. On average, he’ll make about 9 out of 15.
9) σ =
√
npq =
√
15 · 0.60 · 0.40 = 1.90.