Question

Q-3
А
uses
garage
particuda
Selving
spare part at or average rate of
5 per week Assuming that usage
of this spare part follows a poisson
distribution bind the probability that-
a) exactly 5 are used in a particular
Week.
6) at last 5 are used in a particular
weak
used in 3-
c) exactly
15 are used in a 3-week
period.
d) at last 15 are
week period.
e) exactly 5 are used in each of
weak.
3- sed essiv​

Answer 1

Answer:

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Answer 2

Given a poisson distribution with mean of 5 spare parts  use per week, find the given probabilities.

Explanation:

1. The probability mass function of a poisson distribution is,   $P(x)=\frac{\lambda^{x}e^{-\lambda}}{x!}$       -------(a)
2. For a random variable 'x' and λ being the mean which here is $5$ spare parts used per week.  
3. (a) here one week is considered hence probability of exactly $5$ used is given by, $P(x=5)=\frac{5^{5}e^{-5}}{5!}\ \ \ \ \ \ ->P(x=5)=0.175(approx.)$   --ans
4. (b) here three weeks are considered hence the number of spare parts used ranges till $5x3=15$. The probability of at least $5$ used is given by,     [tex]P(x\geq 5)= 1-P(x<5) \\ P(x\geq 5)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)]\\ [/tex]        putting this values in (a) we get , $P(x\geq 5)=1-[0.441] \ \ \ \ \ \ \ \ ->P(x\geq 5)=0.559(approx.)$   --ans    
5. (c) here also three weeks are considered, range of x is till $15$. the probability of exactly $15$ are used is ,                                                                      $P(x=15)= \frac{5^{15}e^{-5}}{15!} \ \ \ \ \ \ \ \ ->P(x=15)=0.000157(approx.)$   ---ans