When a customer places an order with Rudy’s On-Line Office Supplies, a computerized accounting information system (AIS) automatically checks to see if the customer has exceeded his or her credit limit. Past records indicate that the probability of customers exceeding their credit limit is 0.05. Suppose that, on a given day, 20 customers place orders. Assume that the number of customers that the AIS detects are having exceeded their credit limit is distributed as a binomial random variable.
Answers
Explanation:
Given Information:
Total number of customers placed order,
n
=
20
.
Probability of customers exceeding their credit limit,
p
=
0.05
.
Number of customers that the AIS detects as having exceeded their credit limit is distributed as a binomial random variable.
Let X be a binomial random variable that depicts the number of customers that the AIS detects as having exceeded their credit limit. Thus,
X
∼
B
i
n
- o
m
i
a
l
(
n
=
20
,
p
=
0.05
)
The probability mass function of X can be written as:
P
(
X
=
x
)
=
n
C
x
p
x
(
1
−
p
)
n
−
x
=
20
C
x
(
0.05
)
x
(
1
−
0.05
)
20
−
x
a.
For a binomial distribution the mean and standard deviation is given as:
M
e
a
n
,
μ
=
n
×
p
=
20
×
0.05
=
1
S
t
a
n
d
a
r
d
D
e
v
i
a
t
i
o
n
,
σ
=
√
n
×
p
×
(
1
−
p
)
=
√
20
×
0.05
×
0.95
=
0.9747
The mean number of customers exceeding their credit limits is 1.
The standard deviation of the number of customers exceeding their credit limits is 0.9747.
b.
The probability that 0 customers will exceed their limits can be computed as follows:
P
(
X
=
0
)
=
20
C
0
(
0.05
)
0
(
1
−
0.05
)
20
−
0
=
0.3585
Thus, the probability that 0 customers will exceed their limits is 0.3585.
c.
The probability that1 customer will exceed his or her limit is,
P
(
X
=
1
)
=
20
C
1
(
0.05
)
1
(
1
−
0.05
)
20
−
1
=
0.3774
Thus, the probability that1 customer will exceed his or her limit is 0.3774