Chapter 4, Problem 192SE

a.

To determine

Find the percentage for the test needs more than 70 seconds to complete.

Answer to Problem 192SE

The percentage for the test takes 70 seconds and more to complete is 0.2.

Explanation of Solution

Given info:

The probability density function for the time is,

$f\left(x\right)=0.04$ for $50<x<75$.

Calculation:

Let the random variable X represents the time to complete the test in hematology cell counter on a blood sample and the X follows the uniform distribution.

Calculate the probability of test exceeds more than 70 seconds is, $P\left(X>70\right)$.

$\begin{array}{c}P\left(X>70\right)={\displaystyle {\int }_{70}^{75}0.04dx}\\ ={\left(0.04x\right)}_{70}^{75}\\ =\left(0.04\left(75-70\right)\right)\\ =0.04\times 5\\ =0.2\end{array}$

Thus, the value of $P\left(X>70\right)$ is 0.2.

Therefore, the percentage of the completion of the test with more than 70 seconds is 0.2.

b.

To determine

Find the percentage for the test needs less than 1 minute to complete.

Answer to Problem 192SE

The percentage for the test needs less than 1 minute to complete is 0.4.

Explanation of Solution

Calculation:

Convert the minutes into seconds

$1\text{minute}=60\text{ seconds}$

Calculate the probability of the test needs less than 60 seconds is, $P\left(X<60\right)$.

$\begin{array}{c}P\left(X<60\right)={\displaystyle {\int }_{50}^{60}0.04dx}\\ ={\left(0.04x\right)}_{50}^{60}\\ =\left(0.04\left(60-50\right)\right)\\ =0.04\times 10\\ =0.4\end{array}$

Thus, the value of $P\left(X<60\right)$ is 0.4.

Therefore, the percentage of the completion of the test with less than 60 seconds is 0.4.

c.

To determine

Find the mean and variance for the time to compete the test.

Answer to Problem 192SE

The mean and variance for the time to compete the test is 62.5 and 52.0833 respectively.

Explanation of Solution

Calculation:

The formula for mean is,

$E\left(X\right)=\frac{a+b}{2}$

Here the value of a is 50 and b is 75.

The value of mean is,

$\begin{array}{c}E\left(X\right)=\frac{50+75}{2}\\ =\frac{125}{2}\\ =62.5\end{array}$

Thus, the mean value is 62.5 seconds.

The formula for variance is,

$V\left(X\right)=\frac{{\left(b-a\right)}^2}{12}$

The value of variance is,

$\begin{array}{c}V\left(X\right)=\frac{{\left(75-50\right)}^2}{12}\\ =\frac{{\left(25\right)}^2}{12}\\ =\frac{625}{12}\\ =52.0833\end{array}$

Thus, the mean variance is 52.0833 seconds².