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A biostatistician is interested in studying the time X (in seconds) it takes a hematology cell...

Question

A biostatistician is interested in studying the time X (in seconds) it takes a hematology cell...

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 < pi/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 /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.

Answers

Q1) The cumulative distribution fucntion here is obtained as:

F«(x) = P(X S 2) = 1 Cost dt = Sin c

Therefore the CDF here is given as:

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

The median here is computed as:

FM) = 0.5

M = Sin- (0.5) = 6

This is the required median here.

The 65th percentile value is computed in same way here as:

M = Sin-1(0.65) = 0.07177

This is the required 65th percentile value here.

Q2) The percentage of tests that require less than \pi/3 seconds is computed here as:

FT/3) = sin(7/3) 0.8660

Therefore 86.60% is the required percentage here.

Q3) The expected values here are computed as:

-1/2 Elesin 1) = $* (Cos thesin dx = lesin 1177 =e-1

Therefore (e - 1) is the required mean value here.

Now we first compute the second moment here as:

e2sin 1/2 e2-1 E(e2sin 1) = **/2 (Cos x)e2sin I de . = 2 2 0

Now the variance here is computed as:

Var(e^{sin \ x}) = E(e^{2Sinx}) - [E(e^{sinx})]^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}

Var(esin 1) = -0.5e2 – 1.5 + 2e

Var(e^{sin \ x }) = 0.2420

Therefore 0.2420 is the required variance here.


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