STAT 2507
SolutionLab Assignment # 3
Fall 2009
1. Suppose that
X
has a binomial distribution with
n
=25 and
p
=0.8.
Use minitab to
simulate 25 values of
X
.
random 25 c1;
binomial 25 0.8.
(i) [2] How many of your values are less than 21? 18
(ii) [2] How many of your values are between 21 and 24 inclusive? 7.
(iii) If
Y
has a binomial distribution with
n
=23 and
p
=0.8, use the ’cdf’ command, (it
gives you the value of
P
(
Y
≤
k
)),
cdf;
binomial 23 0.8.
to calculate:
P
(
Y <
18) =
P
(
Y
6
17) = 0
.
30531 [2] and
P
(18
≤
Y
≤
22) =
P
(
Y
6
22)

P
(
Y
6
17) = 0
.
9941

0
.
30531 = 0
.
68879 [2]
(iv) If you simulate 10000 values of
Y
, what would be the expected number of values
(among the 10000 values) that are less than 18? 10000(0.3053)=3053 [2]
2. Suppose that
X
has a Poisson distribution with mean
μ
=25. Use the ’cdf’ command
cdf;
poisson 25.
to calculate:
P
(
X <
18) =
P
(
Y
6
17) = 0
.
06048 [2] and
P
(18
≤
X
≤
22) =
P
(
Y
6
22)

P
(
Y
6
17) = 0
.
31735

0
.
06048 = 0
.
25687 [2] The expected number of values
(among the 10000 values) that are less than 18 is 10000(0.06048)=604.8 [2]
3. Suppose that Y has a normal distribution with mean
μ
=35 and variance
σ
2
=4.
To
find
P
(
Y
≤
b
), where
b
is any fixed constant, use the ’cdf’ command
cdf
b
;
normal 35, 2.
To find ’?’ that satisfies
P
(
Y
≤
?) =
a
, where
a
is a fixed know number between 0 and
1, use the ’invcdf’ command
1
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invcdf a;
normal 35, 2.
To simulate 20 observations from the distribution of
Y
, and put them in c1, in minitab:
random 20 c1;
normal 35 2.
(i) Use minitab to find the probabilities
P
(
Y
≤
33) = 0
.
158655 [2],
P
(
Y
>
36) =
1

P
(
Y <
36) = 1

0
.
691462 = 0
.
308538 [2], and
P
(33
≤
Y
≤
36) = 0
.
691462

0
.
158655 = 0
.
532807 [2].
(ii) What is the value of
c
in
P
(
Y
≤
c
)= 0.25? 33.65102. [2]
(iii) Simulate 20 observations from the distribution of Y and use the ’describe’ command
to find the mean, ¯
x
, and the standard deviation
s
for these 20 values. How many of the
values fall in the interval ¯
x
±
2
s
? 19 since ¯
x
±
2
s
= 35
.
374
±
2(1
.
92) = (31
.
534
,
39
.
214)
[4].
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 Fall '09
 MAJIDMOJIRSHEIBANI
 Binomial, Normal Distribution, Standard Deviation, Probability theory, Binomial distribution

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