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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

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#### Solution

Let X denote the number of complaints which a bank manager receives per day.

Given, m = 4 and e^{–4} = 0.0183

∴ X ∼ P(m) = X ∼ p(4)

The p.m.f. of X is given by

P(X = x) = `("e"^-"m" "m"^x)/(x!)`

∴ P(X = x) `("e"^-4(4)^x)/(x!), x` = 0, 1, 2,...

P(only two complaints on a given day)

= P(X = 2)

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

= `(0.0183 xx 16)/(2)`

= 0.1464

#### RELATED QUESTIONS

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**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, .....

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∴ q = `square`, p = `square`, n = 5.

X ~ B`(5, square)`, P(x) = `""^"n""C"_x"P"^x"q"^("n" - x)`

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**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).

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

∴ m = `square`

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

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∴ m = `square`

Now, P[X ≥ 2] = 1 – P[x < 2]

= 1 – {P[X = 0] + P[X = 1]

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**Solution:** Since P(X = 1) = P(X = 2)

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∴ m = `square`

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(Given that e^{-0.2} = 0.8187)

If X ∼ P(m) with P(X = 1) = P(X = 2), then find the mean and P(X = 2).

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Solution: Since P(X = 1) = P(X = 2)

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∴ m = `square`

∴ P(X = 2) = `("e"^-2. 2^2)/(2!)` = `square`