Respuesta :
Answer:
a) P(X ≤ 2) = 0.87
b) P(X ≥ 5) = 0.01
c) P(1 ≤ X ≤ 4) = 0.71
d) P ( X = 0 ) = 0.28
e) σ(X) = 1.09 , E(X) = 1.25
Step-by-step explanation:
Given:
- Let X = the number of defective boards in a random sample of size n = 25, so X ~ Bin(25, 0.05)
Where, n = 25 and p = 0.05
Find:
(a) Determine P(X ≤ 2).(b) Determine P(X ≥ 5).(c) Determine P(1 ≤ X ≤ 4).(d) What is the probability that none of the 25 boards is defective?(e) Calculate the expected value and standard deviation of X.
Solution:
- The probability mass function for a binomial distribution is given by:
P ( X = x ) = nCr * (p)^r * ( 1 - p )^(n-r)
a) P(X ≤ 2):
P(X ≤ 2) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 )
= (0.95)^25 + 25*0.05*0.95^24 + 25C2*0.05^2*0.95^23
= 0.87
b) P(X ≥ 5):
P(X ≥ 5) = 1 - [P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P(X=4)
= 1 - [ 0.87 + 25C3*0.05^3*0.95^22 + 25C4*0.05^4*0.95^21]
= 1 - 0.98994
= 0.01
c) P(1 ≤ X ≤ 4):
P(1 ≤ X ≤ 4) = P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P(X=4)
= ( 0.87 - 0.95^25) + 0.11994
= 0.71
d) P( X = 0 )
P ( X = 0 ) = 0.95^25 = 0.28
e) E(X) & σ(X):
E(X) = n*p
E(X) = 25*0.05 = 1.25
σ(X) = sqrt ( Var (X) )
σ(X) = sqrt ( n*p*(1-p) ) = sqrt ( 25*0.05*0.95 )
σ(X) = 1.09
The answer for the subparts are :
a) P(X ≤ 2) = 0.87
b) P(X ≥ 5) = 0.01
c) P(1 ≤ X ≤ 4) = 0.71
d) P ( X = 0 ) = 0.28
e) σ(X) = 1.09 , E(X) = 1.25
"Standard Deviation"
Given :
Let X = the number of defective boards in a random sample of size
n = 25
p = 0.05
Formula:
P ( X = x ) = [tex]nCr * (p)^r * ( 1 - p )^(n-r)[/tex]
Answer a)
P(X ≤ 2)
P(X ≤ 2) = [tex]P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 )[/tex]
P(X ≤ 2) = [tex](0.95)^25 + 25*0.05*0.95^24 + 25C2*0.05^2*0.95^23[/tex]
P(X ≤ 2) = [tex]0.87[/tex]
The P(X ≤ 2) is 0.87.
Answer b)
P(X ≥ 5)
P(X ≥ 5) =[tex]1 - [P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P(X=4)[/tex]
P(X ≥ 5) = [tex]1 - [ 0.87 + 25C3*0.05^3*0.95^22 + 25C4*0.05^4*0.95^21][/tex]
P(X ≥ 5) =[tex]1 - [ 0.87 + 25C3*0.05^3*0.95^22 + 25C4*0.05^4*0.95^21][/tex]
P(X ≥ 5) = [tex]0.01[/tex]
The P(X ≥ 5) is 0.01.
Answer c)
P(1 ≤ X ≤ 4)
P(1 ≤ X ≤ 4) =[tex]P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P(X=4)[/tex]
P(1 ≤ X ≤ 4) = [tex]( 0.87 - 0.95^25) + 0.11994[/tex]
P(1 ≤ X ≤ 4) = [tex]0.71[/tex]
The P(1 ≤ X ≤ 4) is 0.71.
Answer d)
P( X = 0 )
P ( X = 0 ) = [tex]0.95^25[/tex]
P( X = 0 )= [tex]0.28[/tex]
The probability that none of the 25 boards is defective is 0.28.
Answer e)
The expected value and standard deviation of X is :
Expected value(X)
- Expected value(X) =[tex]n*p[/tex]
- Expected value(X) =[tex]25*0.05[/tex]
- Expected value(X) = [tex]1.25[/tex]
The expected value is 1.25.
Standard deviation
- σ(X) = [tex]sqrt ( Var (X) )[/tex]
- σ(X) = [tex]sqrt ( n*p*(1-p) ) = sqrt ( 25*0.05*0.95)[/tex]
- σ(X) = [tex]1.09[/tex]
The standard deviation of X is 1.09.
Learn more about "Standard Deviation":
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