Chapter 6.1, Problem 25E

(a)

To determine

To Calculate: the probability $\text{P}\left(-1<Y<1\right)$

Answer to Problem 25E

  $\text{P}\left(-1\le Y<1\right)=0.4$

Explanation of Solution

Formula used:

Density curve for the uniform distribution

  $f\left(x\right)=\frac{1}{a-b}$

Calculation:

Interval from between -1 and 4

  $\begin{array}{c}a=-1\\ b=4\end{array}$

Density curve for the uniform distribution

  $\begin{array}{c}f\left(x\right)=\frac{1}{4-\left(-1\right)}\\ =\frac{1}{5}\\ =0.2\end{array}$

  $-1\le x\le 4$

Probability for between -1 and 1

  $\begin{array}{c}\text{P}\left(-1\le Y<1\right)=(1-\left(-1\right)).\frac{1}{5}\\ =\frac{2}{5}\\ =0.4\\ =40\%\end{array}$ I

Probability between -1 and 1 is 0.40 which implies that class will be dismissed in 1 minute about 40% of the time

(b)

To determine

To Calculate: the ${\mu }_Y$ and explain

Answer to Problem 25E

  $\mu =1.5$

Explanation of Solution

Formula used:

Density curve for the uniform distribution

  $f\left(x\right)=\frac{1}{a-b}$

Calculation:

  $\begin{array}{c}a=-1\\ b=4\end{array}$

Density curve for the uniform distribution

  $\begin{array}{c}f\left(x\right)=\frac{1}{4-\left(-1\right)}\\ =\frac{1}{5}\\ =0.2\end{array}$

  $-1\le x\le 4$

  $\begin{array}{c}\mu =\frac{a+b}{2}\\ =\frac{-1+4}{2}\\ =1.5\end{array}$

The uniform distribution is symmetric, so the mean lies exactly in the middle of the distribution. So mean can be getting as the average of the two boundaries.

(c)

To determine

To Calculate: the value o k

Answer to Problem 25E

  $k=2.75$

Explanation of Solution

  $\begin{array}{c}\text{P}\left(Y\ge k\right)=0.25\\ =\left(4-k\right).\frac{1}{5}\\ =\frac{\left(4-k\right)}{5}\\ \frac{\left(4-k\right)}{5}=0.25\\ 4-k=1.25\\ k=4-1.25\\ k=2.75\end{array}$