Chapter 6.1, Problem 26E

(a)

To determine

To Calculate: $\text{P}\left(Y>3.75\right)$

Answer to Problem 26E

  $\text{P}\left(Y>3.75\right)=0.3125$

Explanation of Solution

Formula used:

Density curve for the uniform distribution

  $f\left(x\right)=\frac{1}{a-b}$

Calculation:

Interval between 1 and 5

  $\begin{array}{c}a=1\\ b=5\end{array}$

Density curve for the uniform distribution

  $\begin{array}{c}f\left(x\right)=\frac{1}{5-1}\\ =\frac{1}{4}\\ =0.25\end{array}$

  $1\le x\le 5$

Probability for greater than 3.75

  $\begin{array}{c}\text{P}\left(Y>3.75\right)=(5-3.75).\frac{1}{4}\\ =1.25\times \frac{1}{4}\\ =0.3125\\ =31.25\%\end{array}$ I

Probability is greater than 3.75 which imply that light flashes 3.75 seconds about 31.25% of the time.

(b)

To determine

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

Answer to Problem 26E

  $\mu =3$

Explanation of Solution

Formula used:

Density curve for the uniform distribution

  $f\left(x\right)=\frac{1}{a-b}$

Calculation:

  $\begin{array}{c}a=1\\ b=5\end{array}$

Density curve for the uniform distribution

  $\begin{array}{c}f\left(x\right)=\frac{1}{5-1}\\ =\frac{1}{4}\\ =0.25\end{array}$

  $1\le x\le 5$

  $\begin{array}{c}\mu =\frac{a+b}{2}\\ =\frac{1+5}{2}\\ =3\end{array}$

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 26E

  $k=2.75$

Explanation of Solution

  $\begin{array}{c}\text{P}\left(Y\le k\right)=0.38\\ =\left(k-1\right).\frac{1}{4}\\ =\frac{\left(k-1\right)}{4}\\ \frac{\left(k-1\right)}{4}=0.38\\ k-1=1.52\\ k=1+1.52\\ k=2.52\end{array}$