A large online retailer advertises that 95% of orders placed online will be deli

Question

A large online retailer advertises that 95% of orders placed online will be delivered on-time. We assume this means that for any single randomly-selected online order, there is a 95% probability it will be delivered on-time. Use this information to answer the following questions (round each answer to at least three decimal places).

A.80 online orders are randomly selected (assume these are independent selections, made with replacement). What is the probability that all 80 of the selected orders are delivered on-time?

B.100 online orders are randomly selected (assume these are independent selections, made with replacement). What is the probability that 95 or more of the selected orders are delivered on-time?

C.160 online orders are randomly selected (assume these are independent selections, made with replacement). What is the probability that 95% or more (152 or more) of the selected orders are delivered on-time?

D.One of this retailer’s distribution centers ships 800 orders on a particular day, 760 of which are delivered on-time. A set of 60 orders from this distribution center on this day are selected (without replacement). What is the probability that 95% or more (57 or more) of the selected orders are delivered on-time?

Explanation / Answer

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
a.
X ~ B(80,0.95)
P( X = 80 ) = ( 80 80 ) * ( 0.95^80) * ( 1 - 0.95 )^0
= 0.016515
b.
P( X > 95) = P(X=99) + P(X=98) + P(X=97) + P(X=96) + P(X=100) + P(X=95)
= ( 100 100 ) * 0.95^100 * ( 1- 0.95 ) ^0+ ( 100 99 ) * 0.95^99 * ( 1- 0.95 ) ^1 + ( 100 98 ) * 0.95^98 * ( 1- 0.95 ) ^2 + ( 100 97 ) * 0.95^97 * ( 1- 0.95 ) ^3 + ( 100 96 ) * 0.95^96 * ( 1- 0.95 ) ^4 + ( 100 95) * 0.95^95 (1-0.95)^5

= 0.615999

c.
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 160 * 0.95 = 152
standard deviation ( npq )= 160*0.95*0.05 = 2.7568
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
GREATER THAN EQUAL
P(X < 152) = (152-152)/2.7568
= 0/2.7568= 0
= P ( Z <0) From Standard NOrmal Table
= 0.5
P(X > = 152) = (1 - P(X < 152))
= 1 - 0.5 = 0.5

d.
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 60 * 0.95 = 57
standard deviation ( npq )= 60*0.95*0.05 = 1.6882
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
GREATER THAN EQUAL
P(X < 57) = (57-57)/1.6882
= 0/1.6882= 0
= P ( Z <0) From Standard NOrmal Table
= 0.5
P(X > = 57) = (1 - P(X < 57))
= 1 - 0.5 = 0.5

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