Question

4) A) In a timeshare computer system, the number of teleport requests is 0.2 per millisecond, on average, and follows a Poisson distribution. So the probability that they won't come requests for the next three milliseconds is:

B) Certain resistors are manufactured with a tolerance of ± 10%. If we consider that the real resistance is evenly distributed within this interval, the probability that a resistance with value rated at 1000 Ω have an actual resistance between 990 and 1010 Ω is:

C) The asking price for a certain item is normally distributed with an average of $ 50.00 and deviation standard $ 5.00. Buyers are willing to pay an amount that is also distributed normally with a mean of $ 45.00 and standard deviation of $ 2.50. The probability that the transaction is made is:

Answer 1

a)

Mean/Expected number of events of interest for 3 millisecond: λ = 0.2*3= 0.6

X	P(X)
0	0.5488

b)

here a=   900
b=   1100

P (    990   ≤ X ≤    1010   ) =(x2-x1)/(b-a) =    0.1

c)

µ =    50                              
σ =    5                              
we need to calculate probability for ,                                  
P (   37.5   < X <   52.5   )                  
=P( (37.5-50)/5 < (X-µ)/σ < (52.5-50)/5 )                                  
                                  
P (    -2.500   < Z <    0.500   )                   
= P ( Z <    0.500   ) - P ( Z <   -2.500   ) =    0.6915   -    0.0062   =    0.6853

THANKS

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