Binomial Hypergeometric Poisson Normal Distributions

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Tutorial 3
Binomial Hypergeometric Poisson Normal
Distributions
Hypergeometric (no replacements)
Binomial (replacements)
Population large enough
plt0.05, ngt30
Poisson (events in interval of
time, volume, with known rate)
Normal (X normally
distributed)
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Tutorial 3
Question 1
40 of all sales calls are successful. Five
calls are made on a particular day and the number
of successes X is observed. Write down the pdf
and cdf of X. Binomial. p 0.4 and q 1 p
0.6 5 calls made so n 5 p.d.f. for Binomial
is P(X x) (nCx) px qn-x
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Tutorial 3
Question 1
Binomial. p 0.4 and q 1 p 0.6 5 calls
made so n 5 p.d.f. for Binomial is P(X x)
(nCx) px qn-x
P(X 0) (5C0) 0.40 0.65 0.078 P(X 1)
(5C1) 0.41 0.64 0.259 P(X 3) (5C2) 0.42
0.63 0.346 P(X 3) (5C3) 0.43 0.62 0.23
P(X 4) (5C4) 0.44 0.61 0.077 P(X 5)
(5C5) 0.45 0.60 0.01
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Tutorial 3
Question 1
Binomial. p 0.4 and q 1 p 0.6 5 calls
made so n 5 p.d.f. for Binomial is P(X x)
(nCx) px qn-x
(a) exactly three successes P(X 3) 0.23 (b)
at most 3 successes P(X lt 3)
P(0)P(1)P(2)P(3) 0.913 (c) more than 3
successes. P(X gt 3) 1P(X lt3) 1 -0.913
0.087 (d) State the number of successful calls.
What is the variance? Mean of Binomial np
5 .4 2 successful sales calls are
expected. Variance of the Binomial npq 5 .4
.6 1.2
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Tutorial 3
Question 2

• 25 of people in a large population are
  known to favour the Honest Democratic Party. If
  5 people are drawn from this population at random
  determine the probabilities that
• None favour the HDP.
• All 5 favour the HDP.
• Between 2 and 4 favour the HDP.
• Binomial. p 0.25 q 0.75 n 5,
  P(X x) (nCx) px qn-x
• a) P(X 0) 0.755 0.237
• (b) P(X 5) (5C5) 0.255 0.00097
• (c) P(2 ltX lt 4 ) P(2) P(3) P(4) 0.364

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Tutorial 3
Question 3
A sample of 5 accounts is chosen without
replacement from a set of 20. From past
experience it is known that 20 of the accounts
are in error. What is the probability that not
more than two errors will occur in the sample?
Hypergeometric distribution n 5 N 20 p
.2 q 1 p .8 M Np 20 .2 4. L
Nq 20 .8 16 (4C0) (16C5) (4C1)
(16C4) (4C2) (16C3) 0.968
(20C5) (20C5) (20C5)
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Tutorial 3
Question 4
In a large accounting firm, a sample of 20
accounts from a very large set of 2,000 items are
subject to examination. From previous
experience, it is known that 10 of all accounts
are in error. (a) If not more than two errors are
found in the sample, the financial statements are
accepted. What is the probability that this will
happen?
(b) Calculate the probability that
between 2 and 4 errors will be found in the
sample. Sampling without replacement from a
large population, use Binomial as an
approximation to the Hypergeometric P(X x)
(nCx) px qn-x n 20 N 2,000 p 0.1 q
0.9 (a) P(X lt 2)P(0) P(1) P(2) 0.920
(20C1) (0.1) (0.9)19 (20C2) (0.1)2 (0.9)18
0.677 (b) P(2 ltX lt 4 ) P(2) P(3) P(4)
0.565
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Tutorial 3
Question 5
The error rate in an accounting population
is thought to be .05. Find the probability that
more than 5 errors will occur in a sample of size
100 line items selected for auditing. p0.05
(since p lt 0.05 it is considered small). n100
(ngt30, it is considered large) Use Poisson as
approximation to Binomial. Poisson e-? ?x
? 0.05 100 5,
x! P(X gt 5 ) 1 P(X lt 5 ) 1 (P(X0)
P(X1) P(X2) P(X3) P(X4) P(X5)) 0.3842

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Tutorial 3
Question 6

• A secretary makes 2 errors per page on
  average. What is the probability that he makes
• 4 or more errors on the next page he types?
• no errors on the next page he types?
• Poisson with Mean ? 2
• (a) P(X gt 4) 1 P(X lt 4) 1 P(X lt 3) 1
  (P(X0) P(X1) P(X2) P(X3)) 0.1431
• (b) P(X 0) e-2 20 0.1353
• 0!

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Tutorial 3
Question 7
A soft drinks machine is regulated so that
it discharges an average of 7 ounces per cup.
The amount of drink is normally distributed with
standard deviation equal to 0.5 ounces.

Z (X 7) / 0.5 (a) What fraction of
cups will contain more than 8.0 ounces?
Z (8 7) / 0.5 2. In Z
table P(Z gt 2 ) 0.023
(b) What is the probability that a
cup contains between 6.5 and 7.5 ounces? P
(6.5 lt X lt 7.5) corresponds to P (-1 lt Z lt 1).
In Z table, P(Z gt 1) .15866. Because
distribution is symmetric, P(Z lt -1) .15866.
Subtracting the two tails from 1 gives 1
(.15866 .15866) 0.683 (c) How many cups are
likely to overflow if 8.2 ounce cups are used for
the next 1,000 drinks? P(X gt 8.2) corresponds to
P(Z gt 2.4) 0.0082. 0.82 of the 1000 cups
would overflow. So 8 cups would overflow. (d)
Below what value do we get the smallest 10 of
the drinks? In Z table,
lowest 0.1 corresponds to Z -1.28. X
0.5Z7 0.5(-1.28)7 6.36. Bottom 10
contain less than 6.36 ounces