हिंदी

The Number of Complaints Which a Bank Manager Receives per Day is a Poisson Random Variable with Parameter M = 4. Find the Probability that the Manager Will Receive - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

The number of complaints which a bank manager receives per day is a Poisson random variable with parameter m = 4. Find the probability that the manager will receive -

(a) only two complaints on any given day.

(b) at most two complaints on any given day

[Use e-4 =0.0183]

योग
Advertisements

उत्तर

`P(X=k)=e^(-m) xx m^k/(k!)`

(a) only two complaints on any given day.

k=2, m=4 

`P(X=2)=e^(-4) xx 4^2/2`

`=0.0183 xx 8 `

`=0.1464`
Probability that the manager will recieve only two complaints on any given day = 0.1464

(b) at most two complaints on any given day

`=P(x<=2)`

`=P(x=0)+P(x=1)+P(x=2)`

`= e^(-4) + 4e^(-4) + 4^2   (e^(-4))/(2!)`

`= e^(-4) ( 1 + 4 + 8)`

`= 13 x e^(-4)`

= 13 x 0.0183

= 0.2379

Probab ility that the manager will receive · at most two complaints on any given day = 0.2379.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2014-2015 (March)

APPEARS IN

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

 If a random variable X follows Poisson distribution such that P(X = l) =P(X = 2), then find P(X ≥ 1).  [Use e-2 = 0.1353] 


If X has a Poisson distribution with variance 2, find P (X = 4) 

[Use e-2 = 0.1353] 


If X has a Poisson distribution with variance 2, find 

Mean of X [Use e-2 = 0.1353] 


If X~P(0.5), then find P(X = 3) given e−0.5 = 0.6065.


The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives only two complaints on a given day


The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives a) only two complaints on a given day, b) at most two complaints on a given day. Use e−4 = 0.0183.


It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has exactly 5 rats inclusive. Given e-5  = 0.0067.


It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has between 5 and 7 rats inclusive. Given e−5 = 0.0067.


If E(X) = m and Var(X) = m then X follows ______.


The expected value of the sum of two numbers obtained when two fair dice are rolled is ______.


Choose the correct alternative:

A distance random variable X is said to have the Poisson distribution with parameter m if its p.m.f. is given by P(x) = `("e"^(-"m")"m"^"x")/("x"!)` the condition for m is ______


State whether the following statement is True or False:

X is the number obtained on upper most face when a die is thrown, then E(x) = 3.5


State whether the following statement is True or False:  

A discrete random variable X is said to follow the Poisson distribution with parameter m ≥ 0 if its p.m.f. is given by P(X = x) = `("e"^(-"m")"m"^"x")/"x"`, x = 0, 1, 2, .....


State whether the following statement is True or False:

If n is very large and p is very small then X follows Poisson distribution with n = mp


If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, using the following activity find the value of m.

Solution: X : Follows Poisson distribution

∴ P(X) = `("e"^-"m" "m"^x)/(x!)`, P(X = 1) = 0.4 and P(X = 2) = 0.2

∴ P(X = 1) = `square` P(X = 2).

`("e"^-"m" "m"^x)/(1!) = square ("e"^-"m" "m"^2)/(2!)`,

`"e"^-"m" = square  "e"^-"m" "m"/2`, m ≠ 0

∴ m = `square`


State whether the following statement is true or false:

lf X ∼ P(m) with P(X = 1) = P(X = 2) then m = 1.


In a town, 10 accidents take place in the span of 50 days. Assuming that the number of accidents follows Poisson distribution, find the probability that there will be 3 or more accidents on a day.

(Given that e-0.2 = 0.8187)

Solution:

Here, m = `square` and X − P(m) with parameter m.

The p.m.f. X is:

P(X = x) = `(e^(−m).m^x)/(x!), x = 0, 1, 2,...`

P(X ≥ 3) = 1 − P(X < 3)

= 1 − [`square + square + square`]

= `1 − [(e^(− 0.2)(0.2)^0)/(0!) + (e^(−0.2)(0.2)^1)/(1!) + (e^(−0.2)(0.2)^2)/(2!)]`

= 1 − [0.8187(1 + 0.2 + 0.02)]

= 1 − `square`

= `square`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×