If x is a binomial random variable, compute P(x) for each of the following cases:
(a) P(x≤5),n=7,p=0.3
P(x)=
(b) P(x>6),n=9,p=0.2
P(x)=
(c) P(x<6),n=8,p=0.1
P(x)=
(d) P(x≥5),n=9,p=0.3
P(x)=
If x is a binomial random variable, compute p(x) for each of the
following cases:
(a) n=3,x=2,p=0.9 p(x)=
(b) n=6,x=5,p=0.5 p(x)=
(c) n=3,x=3,p=0.2 p(x)=
(d) n=3,x=0,p=0.7 p(x)=
If x is a binomial random variable, compute ?(?) for each of the
following cases:
(a) ?(?≤1),?=3,?=0.4
?(?)=
(b) ?(?>1),?=4,?=0.2
?(?)=
(c) ?(?<2),?=4,?=0.8
?(?)=
(d) ?(?≥5),?=8,?=0.6
?(?)=
1 point) If xx is a binomial random variable, compute P(x)P(x)
for each of the following cases:
(a) P(x≤1),n=5,p=0.3P(x≤1),n=5,p=0.3
P(x)=P(x)=
(b) P(x>3),n=4,p=0.1P(x>3),n=4,p=0.1
P(x)=P(x)=
(c) P(x<3),n=7,p=0.7P(x<3),n=7,p=0.7
P(x)=P(x)=
If x is a binomial random variable, compute P(x) for each of the
following cases:
(a) P(x≤5),n=9,p=0.7P(x≤5),n=9,p=0.7
(b) P(x>1),n=9,p=0.1P(x>1),n=9,p=0.1
(c) P(x<3),n=5,p=0.6P(x<3),n=5,p=0.6
(d) P(x≥1),n=6,p=0.9P(x≥1),n=6,p=0.9
If x is a binomial random variable, compute the mean, the
standard deviation, and the variance for each of the following
cases:
(a) n=3,p=0.9
μ=
σ^2=
σ=
(b) n=6,p=0.1
μ=
σ^2=
σ=
(c) n=4,p=0.6
μ=
σ^2=
σ=
(d) n=5,p=0.8
μ=
σ^2=
σ=
Suppose that x is a binomial random variable with
n = 5, p = .66, and q = .34.
(b) For each value of x, calculate
p(x). (Round final
answers to 4 decimal places.)
p(0) =
p(1)=
p(2)=
p(3)=
p(4)=
p(5)
(c) Find P(x = 3).
(Round final answer to 4 decimal
places.)
(d) Find P(x ≤ 3).
(Do not round intermediate calculations.
Round final answer to 4 decimal places.)
(e) Find P(x < 3).
(Do not round intermediate calculations....
Determine whether or not the random variable X is a binomial
random
variable.
(a)
X is the number of dots on the top face of a fair die
(b)
X is the number of hearts in a five card hand drawn (without
replacement) from a well
shuffled ordinary deck.
(c)
X is the number of defective parts in a sample of ten randomly
selected parts coming from a manufacturing process in which 0.02%
of all parts are defective.
(d)
X...
The p.d.f of the binomial distribution random variable X with
parameters
n and p is
f(x) =
n
x
p
x
(1 − p)
n−x x = 0, 1, 2, ..., n
0 Otherwise
Show that
a) Pn
x=0 f(x) = 1 [10 Marks]
b) the MGF of X is given by [(1 − p) + pet
]
n
. Hence or otherwise show
that E[X]=np and var(X)=np(1-p).