7.3. Binomial Distribution and Binomial Experiments http://www.ck12.org
a. What is the expected number of households with annual incomes less than $56,400?
b. What is the standard deviation of households with incomes less than $56,400?
c. What is the probability of getting at least 18 out of the 24 households with annual incomes under
$56,400?Review Answers
- (a) The operative word in this problem is ’median’; if the median income is $56,400, then this indicates that
one-half of the income falls below $56,400 and one-half of the income lies above it. Therefore, the chance of
a randomly selected income being below the median income is 0.5. LetXrepresent the number of households
with incomes below the median in a random sample of size 24.Xhas a binomial distribution withn=24 and
p= 0 .5.
E(X) =n p=μx
E(X) = ( 24 )( 0. 5 ) = 12
(b) The standard deviation of households with incomes less than $56,400 is
σx=
√
n p( 1 −p)
σx=√
12 ( 1 − 0. 5 )
σx= 2. 25
(c)P(X≥ 18 )≈
(
n
k)
=
n!
k!(n−k)!(
n
k)
=
n!
k!(n−k)!(
n
k)
=
n!
k!(n−k)!
(
24
18)
=
24!
18!( 24 − 18 )!
(
24
19
)
=
24!
19!( 24 − 19 )!
(
24
20
)
=
24!
20!( 24 − 20 )!
(
24
18
)
= 134596
(
24
19
)
= 42504
(
24
20
)
= 10626
(
n
k)
=
n!
k!(n−k)!(
n
k)
=
n!
k!(n−k)!(
n
k)
=
n!
k!(n−k)!
(
24
21)
=
24!
21!( 24 − 21 )!
(
24
22
)
=
24!
22!( 24 − 22 )!
(
24
23
)
=
24!
23!( 24 − 23 )!
(
24
21
)
= 2024
(
24
22
)
= 276
(
24
23
)
= 24
(
n
k)
=
n!
k!(n−k)!P(X= 18 ) = 134596 ( 0. 5 )^24 ≈ 0. 0080
(
24
24
)
=
24!
24!( 24 − 24 )!
P(X= 19 ) = 42504 ( 0. 5 )^24 ≈ 0. 0025
(
24
24
)
= 1 P(X= 20 ) = 10626 ( 0. 5 )^24 ≈ 0. 0006
P(X= 21 ) = 2024 ( 0. 5 )^24 ≈ 0. 0001
P(X= 22 ) = 276 ( 0. 5 )^24 ≈ 0. 000016
P(X= 23 ) = 24 ( 0. 5 )^24 ≈ 0. 0000014
P(X= 24 ) = 1 ( 0. 5 )^24 ≈ 0. 00000006
The sum of these probabilities gives the answer to the question: 0. 01121746.