Question

a) In a small country, the probability that a person will die from a certain respiratory infection is 0.004. Let ? be the random variable representing the number of persons infected who will die from the infection. A random sample of 2000 persons with this disease is chosen.

(i) Determine the exact distribution of ? and state TWO reasons why it was chosen? [4 marks]

(ii) State the values of ?(?) and ???(?). [2 marks]

(iii) Using a suitable approximate distribution, find the probability that fewer than 5 persons will die from the infection. (Do not use the exact distribution in part (i)). [4 marks]

Answer 1

(i)

The exact distribution of X is binomial with parameters n = 2000 and p = 0.004

i.e

X ~ Binomial (2000,0.004)

Two reasons for choosing binomial distribution for modelling this particular situation:

• There are only two outcomes to this particular scenario : Either death of the patient or survival of the patient from the respiratory infection and the probability of death (success) for each person is constant i.e. 0.004.
• The replications are independent, meaning that the death of one participant due to the respiratory infection does not impact the probability of dying of another participant suffering from that disease.

(ii)

E(X) = n * p = 2000 * 0.004 = 8

Var(X) = n * p * (1-p) = 2000 * 0.004 * (1 - 0.004) = 7.968

(iii)

For large values of n, we use normal approximation to the binomial distribution :

X ~ Binomial (2000,0.004)

tends to

X ~ Normal ( E(X), Var(X) )

i.e.

X ~ Normal( 8 , 7.968 )

We have to find P(X<5)

Without continuity correction*

P(X<5) = P((X-8)/(7.968)^0.5 < (5-8)/(7.968)^0.5)

=P(Z < -1.06279)

= 0.1446

With continuity correction*

P(X<5) = P(X< 5.5) = P((X-8)/(7.968)^0.5 < (5.5-8)/(7.968)^0.5)

=P(Z < -0.88566)

= 0.1867

Here Z = X - 8 / (7.968)^0.5 ~ N (0,1)

* Continuity correction : When we use normal approximation of binomial, a discrete distribution is being approximated by a continous one so we allow for a continuity correction.