Question

If the probability that an individual suffer bad reaction from a certain injection is 0.001 determine the probability that out off 200 individual 1
1.exactly 3
2.more than 2 individuals
3.none
4.more than 1 individual wil suffer a bad reaction

Answer 1

Answer:

1. 3/200

2. 198/200= 99/100

3.0

4. 199/200

Answer 2

Answer:

p ( x > 2 ) = 1 − 5 e 2 = 0.323

Explanation:

The opportunity density characteristic of Poisson’s distribution is p ( x ) = e − λ ⋅ λ x x ! p(x)=e−λ⋅λxx! Where λ = mean = np n = quantity of overall consequences p = opportunity of fulfillment Note: This distribution makes use of in which the opportunity of fulfillment i.e. p may be very small. Calculation: We are given opportunity that on man or woman suffers a horrific response from a positive infection = p = 0.001 n = 2000 (overall no. of individuals) ‘’p’’ may be very small so we are able to use Poisson’s distribution λ = mean = 2000 × 0.001 = 2 ⇒ p ( x ) = e − 2 ⋅ ( 2 ) x x ! ⇒p(x)=e−2⋅(2)xx! Probability of greater than  man or woman suffers ⇒ p(x > 2) = 1 – p (x ≤ 2) And p (x ≤ 2) = p(0) + p(1) + p(2) p(x ≤ 2) = e-2 + 2⋅ e-2 + 2e-2 = 5e-2 ⇒ p ( x > 2 ) = 1 − p ( x ≤ 2 ) = 1 − five e − 2 = 1 − five e 2 ⇒p(x>2)=1−p(x≤2)=1−5e−2=1−5e2 ⇒ p ( x > 2 ) = 1 − five e 2 = 0.323

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