Question

Renal Disease:
The presence of bacteria in a urine sample (bacteriuria) is sometimes associated with symptoms of kidney disease in women. Suppose a determination of bacteriuria has been made over a large population of women at one point in time and 5% of those sampled are positive for bacteriuria.

1. If a sample size of 5 is selected from this population, what is the probability that 1 or more women are positive for bacteriuria?

2. Suppose 100 women from this population are sampled. What is the probability that 3 or more of them are positive for bacteriuria?
One interesting phenomenon of bacteriuria is that there is a “turnover”; that is, if bacteriuria is measured on the same woman at two different points in time, the results are not necessarily the same. Assume that 20% of all women who are bacteriuric at time 0 are again bacteriuric at time 1 (1 year later), whereas only 4.2% of women who were not bacteriuric at time 0 are bacteriuric at time 1. Let X be the random variable representing the number of bacteriuric events over the two time periods for 1 woman and still assume that the probability that a woman will be positive for bacteriuria at any one exam is 5%.

3. what is the probability distribution of X?

4. what is the mean of X?

5. what is the variance of X?

Answer 1

1.

Let X be the number of women are positive for bacteriuria out of sampled 5. Then X ~ Binomial(n = 5, p = 0.05)

probability that 1 or more women are positive for bacteriuria = P(X 1)

= 1 - P(X = 0)

= 1 - ⁵C₀ * 0.05⁰ * (1 - 0.05)^5-0

= 1 - 0.95⁵

= 0.2262191

2.

Let Y be the number of women are positive for bacteriuria out of sampled 100. Then X ~ Binomial(n = 100, p = 0.05)

probability that 3 or more women are positive for bacteriuria = P(Y 3)

= 1 - (P(Y = 0) + P(Y = 1) + P(Y = 2))

= 1 - ( ¹⁰⁰C₀ * 0.05⁰ * (1 - 0.05)^100-0 + ¹⁰⁰C₁ * 0.05¹ * (1 - 0.05)^100-1 + ¹⁰⁰C₂ * 0.05² * (1 - 0.05)^100-2 )

= 1 - (0.005920529 + 0.031160680 + 0.081181772)

= 0.881737

3.

X be the random variable representing the number of bacteriuric events over the two time periods for 1 woman

So, the value of X van be 0 (woman will not be positive at both times) , 1 (woman will be positive at one of the times) , 2 (woman will be positive at both times)

The probability distribution of X is,

P(X = 0) = probability that a woman will not be positive for bacteriuria at time 1 * probability that a woman will not be positive for bacteriuria at time 2 given that woman will not be positive for bacteriuria at time 1

= (1 - 0.05) * (1 - 0.042) = 0.9101

P(X = 1) = probability that a woman will not be positive for bacteriuria at time 1 * probability that a woman will be positive for bacteriuria at time 2 given that woman will not be positive for bacteriuria at time 1

+ probability that a woman will be positive for bacteriuria at time 1 * probability that a woman will not be positive for bacteriuria at time 2 given that woman will be positive for bacteriuria at time 1

= (1 - 0.05) * 0.042 + 0.05 * (1 - 0.20)  = 0.0799

P(X = 2) = probability that a woman will be positive for bacteriuria at time 1 * probability that a woman will be positive for bacteriuria at time 2 given that woman will be positive for bacteriuria at time 1

= 0.05 * 0.2  = 0.01

The probability distribution of X is,

P(X = 0) = 0.9101

P(X = 1) = 0.0799

P(X = 2) = 0.01

4.

Mean of X = E(X) = 0 * 0.9101 + 1 * 0.0799 + 2 * 0.01 = 0.0999

5.

E(X²) = 0² * 0.9101 + 1² * 0.0799 + 2² * 0.01 = 0.1199

variance of X = E(X²) - [E(X)]² = 0.1199 - 0.0999² = 0.10992