Please solve 4 parts. Thank you! A. What is the probability that all three order
ID: 3362859 • Letter: P
Question
Please solve 4 parts. Thank you!A. What is the probability that all three orders will be filled correctly?
B. What is the probability that none of the three orders will be filled correctly?
C. What is the probability that at least two weeks of the three orders will be filled correctly?
D. What are the mean and standard deviation of the bionomial distribution used in (a) through (c)? Interpret these values.
Homework: Week 3 Homework Save core: 0.17 of 1 pt 9of12(12complete) Hw Score: 61.04%, 7.32 of 12 pts 5.2.17 ,E Question Help * Suppose that you and two friends go to a restaurant, which last month filed approximately 908% of the orders correctly. Complete parts a through d) below. a. What is the probability that all three orders will be filled correctly? The probability is Round to four decimal places as needed.) Enter your answer in the answer box and then click Check Answer. 5 Pemaining Clear All Check Answer
Explanation / Answer
5.2.17.
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 = 3 * 0.908
= 2.724
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 3 * 0.908 * 0.092
= 0.2506
III.
standard deviation = sqrt( variance ) = sqrt(0.2506)
=0.5006
a.
probability that all three orders will be filled correctly
P( X = 3 ) = ( 3 3 ) * ( 0.908^3) * ( 1 - 0.908 )^0
= 0.7486
b.
the probability that none of the three orders will be filled correctly
P( X = 0 ) = ( 3 0 ) * ( 0.908^0) * ( 1 - 0.908 )^3
= 0.0008
c.
the probability that at least two weeks of the three orders will be filled correctly
P( X < = 0.2142) = P(X=0.2142)
= ( 3 0.2142 ) * 0.908^0.2142 * ( 1- 0.908 ) ^2.7858 + ( 3 -0.7858 ) * 0.908^-0.7858 * ( 1- 0.908 ) ^3.7858
= 0.0008
P( X > = 0.2142) = 1 - P ( X <=0.2142) = 1 -0.0008 = 0.9992
d.
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 3 * 0.908
= 2.724
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 3 * 0.908 * 0.092
= 0.2506
III.
standard deviation = sqrt( variance ) = sqrt(0.2506)
=0.5006