Stat 2300--Chapter 5 Calculator Helps
TI-83/84
Combinations
• Locate the MATH key
Calculator Helps
Chapter 5
Binomial and Poisson Distributions
Combinations
• Example: To find
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7
MATH
PRB
3:nCr ENTER
3
ENTER
⎛7⎞
⎜3⎟
⎝ ⎠
, type
Binomial Distribution Functions
• Page down to 0:binompdf(
• The syntax is
binompdf(n,p,x)
• This gives the probability of x
successes in n trials where p
is the probability of a
success.
• Press > until PRB is
highlighted.
The DISTR Menu
• We will use the DISTR key
to access the distribution
functions of the calculator.
• Locate the DISTR key on
your calculator.
• Press 2ND VARS to
access DISTR.
Example
• Exercise 34 on page 208
• Forty percent of business travelers
carry either a cell phone or a laptop.
For a sample of 15 business travelers,
make the following calculations.
• (a) Compute the probability that 3 of the
travelers carry a cell phone or a laptop.
• (b) Compute the probability that 12 of the
travelers carry neither a cell phone nor a
laptop.
• (c) Compute the probability that at least
three of the travelers carry a cell phone or a
laptop.
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Stat 2300--Chapter 5 Calculator Helps
TI-83/84
Part (a)
• Here, n=15, p=.40, and x=3.
• Type
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DISTR
0:binompdf(
15
, (the comma is above the number 7)
.4
,
3
) (the final parenthesis is optional)
ENTER
Note that this rounds to
.0634, which is the
number on the binomial
table on page 942.
Part (b)
• This part of the problem is asking about the
opposite event, travelers who carry neither a cell
phone nor a laptop.
• The probability of “success” in this part is 1.40=.60
• So, n=15, p=.6, and x=12.
• Note that this is the same answer we get in part
(a). Why is that? (Hint: If exactly 12 travelers have neither a cell phone
nor a laptop, what must the other 3 have?)
Part (c)
•
Compute the probability that at least three of the travelers carry a
cell phone or a laptop.
• “At least three” means three or more. The
easiest way to calculate this is to find the
probability of getting 2 or fewer and
subtract it from 1.
• In mathematical notation:
• P(at least 3) = 1 – P(2 or fewer)
• = 1 – [f(0) + f(1) + f(2)]
• The binomcdf function lets us compute
f(0) + f(1) + f(2) all at once. It is called a
cumulative density function.
Poisson Distributions
• Exercise 41 on page 212 says:
• During the period of time that a local university
takes phone-in registrations, calls come in at
the rate of one every two minutes.
• (a) What is the expected number of calls in one
hour?
• (b) What is the probability of three calls in five
minutes?
• (c) What is the probability of no calls in a five-minute
period?
Part (c), continued
• Choose A:binomcdf(
• Syntax is binomcdf(n,p,x)
• This gives the probability of
x or fewer successes in n
trials with probability of
success p.
• Enter binomcdf(15, .4, 2)
• Subtract this result from 1.
• There is a .9729 probability
that at least 3 travelers
carry a cell or a laptop.
Part (b)
• We need to find the expected number
of calls in 5 minutes.
⎛ 1call ⎞
⎜
⎟ × ( 5min ) = 2.5calls = μ
⎝ 2min ⎠
• Use B:poissonpdf(
• Part (a) does not require a probability
density function.
⎛ 1 call ⎞ ⎛ 60min ⎞ 30 calls
⎜ 2min ⎟ × ⎜ 1 hour ⎟ = hour
⎝
⎠ ⎝
⎠
• Syntax is poissonpdf(µ, x)
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Stat 2300--Chapter 5 Calculator Helps
TI-83/84
Part (b), continued
• Type
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DISTR
B:poissonpdf(
2.5
,
3
)
• The probability of 3 calls
in 5 minutes is .2138.
Part (c)
• Here we want the
probability of no calls in 5
minutes. Again, µ=2.5
and now x=0.
• The probability is .0821.
• Note: the poissoncdf
function can be used in a
similar manner to the
binomcdf function.
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