Solutions to some problems on binomial distribution
Solutions to some problems on binomial distribution
( is same as p (book uses ( and I used p in place)
7.50
Each birth of a puppy results in a male puppy or female puppy
X = # number o female puppies in a litter of 5
That is n=5
What is p? or (what is ( ?) where p =P( that the puppy is a female)
Since both outcomes (male or female) are equally likely, this implies that p =0.5 ( or ( =0.5)
The distribution can be listed from the table given in the text books
|x |p(x) |
|0 |.0313 |
|1 |.1563 |
|2 |.3125 |
|3 |.3125 |
|4 |.1563 |
|5 |.0313 |
7.51
n= number of trustworthy people
x = # trustworthy people that fail the polygraph test.
p = P( a trustworthy person fails the polygraph test) (=()
It is given in the description that such a p = .15 (=()
Here x is a binomial variable with n=10 and p=.15
a. P( all ten pass) = P( none one fails) = P( x=0) = .1968744 ( a TI calculator gives this value)
b. P(more than 2 fail) = P( x>2) = P(x =3 or x=3 or…..or x=10)
= 1- P(x1
A lot will be accepted if x ≤1
a. Given that whole population contains 5% defective parts
It is given that p = P( a randomly selected part being defective ) = .05
To compute P(the lot being accepted) = P(x ≤1) = P(x=0) +P(x=1)
From table = .358+.377
= -------------
b. Given that whole population contains 10% defective parts
It is given that p = P( a randomly selected part being defective ) = .1
To compute P(the lot being accepted) = P(x ≤1) = P(x=0) +P(x=1)
From table = .122+.058
= ------------
c. Given that whole population contains 20% defective parts
It is given that p = P( a randomly selected part being defective ) = .2
To compute P(the lot being accepted) = P(x ≤1) = P(x=0) +P(x=1)
From table = .012+.058
= ---------------
7.55
From the whole population of automobiles, 30% fail emission test.
That is P( a randomly selected automobile fails the test) = p =.30
X = # cars that fail the inspection out of chosen n cars.
a. If we choose n= 15 cars then to compute P(x≤5) = P(x=0 or x=1 or ….or x=5)
=P(x=0)+…+P(x=5)
= ___+___+___+___+___+___ (look up from the table the find the answer)
b. n = 15 and p = .3 compute P(5≤x≤10) = P(x=5 or x=6 or …or x=10)
= P(x=5)+…+P(x=10)
= ___+___+___+___+___+___ (look up from the table the find the answer)
c. n =25 and p = P( a car passes the inspection) = 1-.30 = .70
( = n*p = 25*.70 = 17.5
Interpretation : if we repeatedly test the samples of size 25 then on the average we will find 17.5 cars/25 cars pass the test.
Standard deviation = σ = (n*p*(1-p) = 2.29128 cars/25cars.
d. n =25
p = .70 ( that car passes the test)
within 1-sd of the mean value means x lies between (-σ and (+σ
P ((-σ≤ x≤ (+σ) = P((5.6930≤ x≤ 10.2756) = P(x =6 or ….or x=10)
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