There are 32 students in a class. 14 have already had a course like this one eit
ID: 3047479 • Letter: T
Question
There are 32 students in a class. 14 have already had a course like this one either in high school or in college, the other have not. if I randomly pick ten students to work in a group. Find the following probabilities 1. a. That all 10 have had a similar class in the past b. That at most 2 have had a similar class in the past A quality control office of the company tests 17 products a day. The company says that the defective rate of their product is 15%. what is the probability that they will have at least 15defective products 2.Explanation / Answer
Q1.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 10 * 0.4375 = 4.375
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability, q = failure probability
variance = 10 * 0.4375 * 0.5625
= 2.460938
III.
standard deviation = sqrt( variance ) = sqrt(2.460938) =1.568738
N=10, P = no.of who had a past = 14/32 = 0.4375
a.
P( X = 10 ) = ( 10 10 ) * ( 0.4375^10) * ( 1 - 0.4375 )^0
= 0.000257
b.
P( X < = 2) = P(X=2) + P(X=1) + P(X=0)
= ( 10 2 ) * 0.4375^2 * ( 1- 0.4375 ) ^8 + ( 10 1 ) * 0.4375^1 * ( 1- 0.4375 ) ^9 + ( 10 0 ) * 0.4375^0 * ( 1- 0.4375 ) ^10
= 0.114164
Q2.
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 17 * 0.15
= 2.55
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 17 * 0.15 * 0.85
= 2.1675
III.
standard deviation = sqrt( variance ) = sqrt(2.1675)
=1.472243
P(atleast 15 defective in product) = P( X > = 15) = P(X=17) + P(X=16) + P(X=15)
= ( 17 17 ) * 0.15^17 * ( 1- 0.15 ) ^0 + ( 17 16 ) * 0.15^16 * ( 1- 0.15 ) ^1 + ( 17 15 ) * 0.15^15 * ( 1- 0.15 ) ^2
P( X = 15 ) = ( 17 15 ) * ( 0.15^15) * ( 1 - 0.15 )^2
= 0.000000000043027
P( X = 16 ) = ( 17 16 ) * ( 0.15^16) * ( 1 - 0.15 )^1
= 0.000000000000949
P( X = 17 ) = ( 17 17 ) * ( 0.15^17) * ( 1 - 0.15 )^0
= 0.00000000000001
P( X > = 15) = 0.000000000043027 + 0.000000000000949 + 0.00000000000001 = 0.000000000043986
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.