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escapelife:
In order to be offered a scholarship, a candidate has to pass two rounds of interview. (It is assumed that all interviewers’ decisions are independent.)

In the first round, there will be a panel of 10 interviewers and the probability of each interviewer passing a candidate is 0.9. The candidate fails to qualify for the second round if more than one interviewer decides not to pass him or her.

(i)   Find the probability that a candidate passes the first round of interview.   

In the second round, there will be a panel of 5 interviewers and the probability of each interviewer passing a candidate is 0.8. The candidate is offered a scholarship only if all interviewers pass him or her in the second round.

(ii)   Show that the probability that the candidate is offered the scholarship is 0.241, correct to three significant figures.            
                                                                               
(iii)   There are n candidates going for the interviews. Find the smallest n such that there is at least a 98% chance of 2 or more candidates being offered the scholarships.

   The panels interview 5 candidates per day for a period of 55 days. Find the probability that, on average, at least 1 candidate is offered the scholarship per day out of this period of 55 days.       



10.   The lifespan of a halogen bulb is normally distributed with mean 160 hours with standard deviation 10 hours, while the lifespan of a fluorescent bulb is normally distributed with mean 240 hours and standard deviation 12 hours. The lifespan of any bulb is independent of one another.

   (i)   A halogen bulb is randomly chosen. Find the greatest value of a, correct to three significant figures, if the probability that its lifespan lies in the range of (160-a, 160+a) is at most 0.4.   
   (ii)   Find the probability that the difference between the average lifespan of two fluorescent bulbs and twice the lifespan of a halogen bulb does not exceed 70 hours.   
   
   (iii)   The halogen bulbs are packed in boxes of n bulbs, where n is large. If there are more than 10 bulbs that have lifespans of less than 150 hours, the box will be rejected.  Using a suitable approximation, find the greatest value of  n so that the probability that a box will be rejected is less than 0.2.


11.    In a small company, the employees send an average of 1.2 print jobs to the colour printer and ? print jobs to the laser printer per day. It is assumed that these are the only two printers in the company and the print jobs are independent.

(i)   Given that on 1 in 100 working days there are no print jobs for both printers.  Show that  = 3.41 correct to 3 significant figures. 
(ii)   Find the probability that a total of 3 print jobs were sent in on a working day.   

(iii)   A typical working day consists of 8 hours of work.  Find the probability that more than half of the total print jobs sent during a typical working day occurs within the first hour of work, given that there was a total of 3 print jobs for the day.         

sm:
 ??? this is not pure math, its prob. & statistics

escapelife:
Ya my teacher says this are questions for my Maths in A level =_=. FYI from Singapore

louis:
Escapelife, i will answer  your first question:

Let X =number of interviewers who will accept the candidate in the first round.
X¬B ( 10,  0.9 )
P(X>=9)
=P(X=9,10)
=(10)*(0.9^9)*(0.1)  +  (0.9^10)
=0.3874 +  0.3487
=0.7361

For each candidate, Probability of qualifying for scholarship
=0.7361 x ( 0.8^5)
=0.241


Let A be the number of candidates who are qualified for the scholarship from a total of n candidates

A¬ B ( n, 0.241 )
P(A>=2) >=0.98

1- P(A=0) -P(A=1)  >= 0.98

Let P(n) = 1 - (0.759)^n  -  n(0.241)(0,759)^(n-1)
Using the trial method,
n=20, p(n) =0.9704
n=21,p(n)  =0.9766
n=22,p(n)  =0.9814
The least number of n = 22

Let Y be the number of candidates accepted from the five candidates for one day

Y ¬B(5, 0.241 )
The number of days n = 55
n is larger i.e. > 30.
The binomial distribution can be approximated to normal distribution by CLT,

Let the average number of candidates accepted per day out of 55 days be R,
 mean = 5x 0.241 = 1.205
 Variance = np(1-p) /  n

             = 5 x 0.241 x0.759  / 55
             = 0.016629
 R ¬ N ( 1.205,  0.016629 )

P(R>=1) =P{ Z > (1-1.205)/ sqrt 0.016629  }

            =P{-1.5897)
            =0.944

louis:
Answer for Second question

Let H and F denote the lifespan of a halogen and fluorescent bulb.

H ~N ( 160, 100 )
F ~N (  240, 144 )

P ( 160 -a < H < 160 + a ) <= 0.4
P ( Z > a ) = 0.3
looking at the Normal Distribution Table,
           a = 5.24

Let 2 Fluorescent bulb be F1 and F2

The average lifespan of 2 fluorescent bulbs = ( F1 + F2 ) / 2
Twice the life span of a halogen bulb be  2 H
 Let ( F1 + F2 ) / 2 - 2 H = D

P{ l D l  < = 70 }
 =P{ -70 <= D <=  70 }
 =0.323

Let X be the number of bulbs from n halogen bulbs that have lifespan less than 150 hrs.

X ~B {  n, P(H<150) }
P(H<150) = P{Z < (150 -160) / 10 )
             = p ( Z < -1)
             = 0.1587

Since x ia large, np>5,nq>5,  x can be approximated to normal distribution,
x~N(0.1587n, 0.1587 x 0.8413 n )
x~P(0.1587n, 0.1335n  )

P(X>10 ) = P {x > 10.5 } < 0.2  ( c.c.)
P{Z > (10.5 -0.1587n)/ sqrt ( 0.1335n) }  <  0.2

(10.5 -0.1587n)/ sqrt ( 0.1335n) } > o.8416
n < 52.181

Greatest n = 52



 

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