Question

A teaching hospital in South-West Part of Nigeria receives on the average 5 pregnant women with high blood pressure per week. What is the probability that on a particular week, the teaching hospital will receive:

1.) No high BP pregnant woman

Answer 1

The probability that on a particular week, the hospital will receive on high BP pregnant woman is 0.0068

Explanation:

We use the Exponential distribution,

Since we are given that on average, 5 pregnant women with high blood pressure come per week,

So, average = m = 5

Now, on average, 5 people come every week, so,

5 women per week,

so, we get 1 woman per (1/5)th week,

Hence, the mean is m = 1/5 for a woman arriving

and λ = 1/m = 5 = λ

we have to find the probability that it takes higher than a week for a high BP pregnant woman to arrive, i.e,

P(X>1) i.e. the probability that it takes more than a week for a high BP pregnant woman to show up,

Now,

P(X>1) = 1 - P(X<1),

Now, the probability density function is,

`f(x) = \lambda e^(-\lambda x)`

And the cumulative distribution function (CDF) is,

`CDF = 1 - e^(-\lambda x)`

Now, CDF gives the probability of an event occuring within a given time,

so, for 1 week, we have x = 1, and λ = 5, which gives,

P(X<1) = CDF,

so,

`P(X < 1)=CDF = 1 - e^(-\lambda x)\\P(X < 1)=1-e^(-5(1))\\P(X < 1)=1-e^(-5)\\P(X < 1) = 1 - 6.738*10^(-3)\\P(X < 1) = 0.9932\\And,\\P(X > 1) = 1 - 0.9932\\P(X > 1) = 6.8*10^(-3)\\P(X > 1) = 0.0068`

So, the probability that on a particular week, the hospital will receive on high BP pregnant woman is 0.0068