Question

A biostatistician is interested in studying the time X (in seconds) it takes a hematology cell counter to complete a test on a blood sample. The probability density function of the aforementioned time is: f(x) = ( cos x if 0 < x < π/2 0 elsewhere. 1. Find the cumulative distribution function and compute the median and 65-th percentile of the time to complete a test on a sample. 2. What is the percentage of tests require less than π/3 seconds to complete? 3. Suppose that this biostatistician is interested in the following composite function of time to complete a test on a sample: exp(sin x). Compute the mean and variance of the underlying random variable.

Answer 1

We are given the PDF for X here as:

f(x) = Cos (x) where X ranges from 0 to

TT/2

Q1) The CDF for X here is obtained as:

$F_x(x) = \int_{0}^{x} Cos \ t \ dt = Sin \ x$

$F_x(x) = Sin \ x, 0 < x < \pi/2$

This is the required cumulative distribution function here.

The median here is computed as:

FM) = 0.5

Sin(M) = 0.5

$M = Sin^{-1} \ 0.5 = \frac{\pi}{6}$

This is the required median here.

similarly, the 65th percentile value here is computed as:

$= Sin^{-1} \ 0.65 = 0.0717\pi$

This is the required 65th percentile value here.

b) The percentage of tests that require less than π/3 seconds to complete is computed using the CDF for X here as:

$F_x(x) = Sin \ x, 0 < x < \pi/2$

$F_x(\pi/3) = Sin \ (\pi/3) = 0.8660$

Therefore 86.60% is the required percentage here.

c) The expected value here is computed as:

$E(e^{sin \ x}) = \int_{0}^{\pi/2} Cos \ x e^{sin \ x} \ dx = \left |e^{sin \ x} \right |_{0}^{\pi/2} = e - 1$

Therefore (e - 1) is the required mean value of the given random variable here.

The second moment of the given random variable here is computed as:

$E(e^{2sin \ x}) = \int_{0}^{\pi/2} Cos \ x e^{2sin \ x} \ dx = \left |\frac{e^{2sin \ x}}{2} \right |_{0}^{\pi/2} = \frac{e^2 - 1}{2}$

Therefore the variance here is computed as:

$Var(e^{sin \ x }) = E(e^{2sin \ x}) - E(e^{sin \ x})^2= \frac{e^2 - 1}{2} - (e - 1)^2$

$Var(e^{sin \ x }) = \frac{e^2 - 1 - 2e^2 - 2 + 4e }{2}$

$Var(e^{sin \ x }) = \frac{ -e^2 - 3 + 4e }{2}$

This is the required variance here.