# Shape of a Binomial Distribution Mean, Variance, and SD

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```							MA131 Introductory Statistics
IT133:                                                       TIHE 2005                                                   Lecture
Lecture 8.2 8.2

Shape of a Binomial Distribution

£ right skewed if p < 0.5
£ symmetric if p = 0.5
£ left skewed if p > 0.5

Mean, Variance, and SD of a Binomial Distribution

• Mean        µ = np
σ 2 = npq
• Variance
√
• SD σ = npq

Example 1 A coin is tossed four times. Find the mean, variance, and standard deviation of the number
of heads that will be obtained.
1
Solution.    n = 4, p =   2   =q
1
µ = np = 4                2       =2
2                         1       1
σ = npq = 4                   2       2    =1
√
σ =   npq = 1

Alternatively: This problem can be solved using expected value formula.

No. of heads, X            0            1       2     3    4
1        4           6     4    1
Probability, P (X)          16       16          16    16   16

µ = E(X) =               XP (X)
1            4                              1
= 0              +1             + ··· + 4                  = 2,
16           16                             16

σ2 =        X 2 P (X) − µ2
1                4                               1
= 02             + 12             + · · · + 42                − 22 = 1,
16               16                             16
and σ = 1.     ///

Example 2 A ﬁsherman ﬁnds that approximately 17% of all his ﬁsh go bad by the time he takes them to
the market. The ﬁsherman catches 1,000 ﬁsh.

(a) About how many will go bad by the time he takes them to the market?
Solution.       n = 1000, p = 0.17
µ = np = 1000(0.17) = 170.
Approximately 170 ﬁsh will go bad.

c 2004, RSHavea, MaCS, USP                                       1                                      File updated: September 13, 2004
MA131                                                                                                             Lecture 8.2

(b) Find the standard deviation.
√
Solution. σ = npq = 1000(0.17)(0.83) = 11.88                        ///

Multinomial distribution

This is used if each trial in an experiment has more than two outcomes.

If X consists of events
E1 , E 2 , E 3 , . . . , E k
which have corresponding probabilities p1 , p2 , p3 , . . . , pk of occurring, and

X1 , X 2 , X 3 , . . . , X k

are the respective numbers of times E1 , E2 , . . . , Ek will occur.

The probability that X will occur is
n!
P (X) =                          pX1 pX2 · · · pXk
X1 !X2 !X2 ! · · · Xk ! 1 2           k

where
n = X1 + X2 + X3 + · · · + Xk
and
1 = p1 + p2 + p 3 + · · · + pk .

Example 3 In a large city,

50% choose a movie,
30% choose dinner and a play,
20% choose shopping

as leisure activity. If a sample of ﬁve people is randomly selected, ﬁnd the probability that three are planning
to go to a movie, one to a play, and one to a shopping mall.

Solution.   In this case we have n = 5,

X1 = 3, X2 = 1, X3 = 1,

p1 = 0.50, p2 = 0.30, and p3 = 0.20.
Let X be the event under consideration. Then it follows that
5!
P (X) =          (0.50)3 (0.30)1 (0.20)1 = 0.15         ///
3!1!1!
Example 4 In a music store, a manager found that the probabilities that a person buys zero, one, or two
or more CDs are 0.3, 0.6, and 0.1, respectively. If six customers enter the store, ﬁnd the probability that
one won’t buy any CDs, three will buy one CD, and two will buy two or more CDs.

c 2004, RSHavea, MaCS, USP                                    2                                File updated: September 13, 2004
MA131                                                                                               Lecture 8.2

Solution.   n = 6, X1 = 1, X2 = 3, X3 = 2, p1 = 0.3, p2 = 0.6, and p3 = 0.1. Then
6!
P (X) =            (0.3)1 (0.6)3 (0.1)2
1!3!2!

= 60(0.3)(0.216)(0.01)

= 0.03888      ///

Multinomial distribution

£ has the advantage of allowing one to calculate probabilities when there are more than two outcomes
for each trial in an experiment;

£ is a generalization of binomial distribution.

Poisson Distribution

• Often used to model the frequency with which a speciﬁed event occurs during a particular period of
time;

• Useful when n is large and p is small and when the independent variable occurs over a period of time.

• Provides a good approximation to the distribution of many variables observed in nature, such as the
number of

– deaths in a given region due to a rare disease;
– daily telephone calls received at a switchboard;
– accidents in a manufacturing plant per month;
– air collisions per month;

Probability distribution formula:
e−λ λX
P (X; λ) = P (X = x) =            ,
X!
where λ is a positive real number, x = 0, 1, 2, . . . , and e ≈ 2.7183.

The random variable X is called a Poisson random variable and is said to have the Poisson distribu-
tion with parameter λ.

A Poisson random variable has inﬁnitely many possible values—viz. all nonnegative integers.

Example 5 If there are 200 typographical errors randomly distributed in a 500-page manuscript, ﬁnd the
probability that a given page contains exactly three errors.

Solution.   On average, the number of errors in each page is
200
λ=       = 0.4.
500

c 2004, RSHavea, MaCS, USP                               3                       File updated: September 13, 2004
MA131                                                                                                     Lecture 8.2

With X = 3,
e−λ λX    (2.7183)−0.4 (0.4)3
P (3; 0.4) =     =                     = 0.0072.
X!              3!
Thus there is less than a 1% probability that any given page will contain exactly three errors.             ///

Our next example uses the Table of Poisson Probabilities.

Example 6 A sales ﬁrm receives, on the average, three calls per hour on its toll–free number. For any
given hour, ﬁnd the probability that it will receive the following.

In any given case, λ = 3.

(a) At most three calls.
Solution.    X = 0, 1, 2, 3

P (0; 3) + P (1; 3) + P (2; 3) + P (3; 3) = 0.0498 + 0.1494 + 0.2240 + 0.2240 = 0.6472        ///

(b) At least three calls.
Solution.    The required probability is

1 − [P (0; 3) + P (1; 3) + P (2; 3)] = 1 − 0.4232 = 0.5768.     ///

(c) Five or more calls
Solution.    The required probability is

1 − [P (0; 3) + P (1; 3) + P (2; 3) + P (3; 3) + P (4; 3)] = 0.1848.     ///

Mean and SD of the Poisson probability distribution
√
µ = λ = σ2,             σ=       λ.

Example 7 Let X denote the number of patients arriving at the ER of CWM Hospital between 6:00 pm
and 7:00 pm with parameter λ = 6.9.

(a) Determine and interpret the mean and random variable X.
Solution.    µ = λ = 6.9.
On average, 6.9 patients arrive at the ER between 6:00 PM and 7:00 PM.

(b) Determine the standard deviation of X.
√    √
Solution. σ = λ = 6.9 = 2.6. ///

Study the following distributions for the test.

• Binomial

• Multinomial

• Poisson

c 2004, RSHavea, MaCS, USP                                4                            File updated: September 13, 2004

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