A company receives 60% of its orders over the Internet. a) With a collection of

Question

A company receives 60% of its orders over the Internet. a) With a collection of 18 independently placed orders, what is the probability that between 8 and 10 (including 8 and 10) of the orders are received over the internet? b) What is the probability that at least 9 orders are received over the internet? c) What are the mean and variance for orders received over the internet Twenty players compete in a tournament. a) In how many ways can rankings be assigned to the top five competitors (where winners are assigned 1", 2nd, 3rd, etc)? b) In how many ways can the best five competitors be chosen (without being in any order)?

Explanation / Answer

(8) here we use binomial distribution with parameter, n=18 and p=0.6 and for

Binomial distribution ,P(X=r)=nCrpr(1-p)n-r  

(a)P(8<=X<=10)=P(X=8)+P(X=9)+P(X=10)=0.0771+0.1284+0.1734=0.3789

P(X=8)=0.0771 ( using ms-excel command =BINOMDIST(8,18,0.6,0) )

(8b)P(X<=9)=1-P(X<=8)=1-0.1347=0.8653

P(X<=8)=0.1347(  using ms-excel command =BINOMDIST(8,18,0.6,1) )

(8c)mean=np=18*0.6=11.4 and variance=np(1-p)=18*0.6*(1-0.6)=4.32

(9a) first position can be taken by 20 players

second position can be taken by remaining 19 players

third position can be taken by remaining 18 players

fourth position can be taken by remaining 17 players

fifth position can be taken by remaining 16 players

number of ways of top five competitiors=20*19*18*17*16=1860480

(9b) required number of ways=20C5=15504

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