Class is over! Mr. Shrager does not always let his statistics class out on time. In fact, he

seems to end class according to his own “internal clock.” The density curve here models

the distribution of Y, the amount of time after class ends (in minutes) when Mr. Shrager

dismisses the class on a randomly selected day. (A negative value indicates he ended class

early.)

a) Find and interpret $P(-1\le Y\le 1).$

b) What is $\mu Y$ ? Explain your answer.

c)Find the value of k that makes this statement true: localid="1654015283453" $P(Y\ge k)=0.25$

The density curve here models the distribution of Y, the amount of time after class ends (in minutes) when Mr. Shrager dismisses the class on a randomly selected day.

The distribution is modeled by a uniform distribution on the interval from $-1$minutes to $4$minutes.

The density curve of distribution is

$$\begin{gathered} f\left(x\right)=\frac{1}{b-a} \\ =\frac{1}{4-(-1)} \\ =\frac{1}{5} \\ =0.2 \end{gathered}$$

The probability

$$\begin{gathered} P(-1\le Y\le 1)=1-(-1)\times \frac{1}{5} \\ =\left(1+1\right)\times \frac{1}{5} \\ =\frac{2}{5} \\ =0.40r40\% \end{gathered}$$

The uniform distribution is perfectly symmetric, which implies that the mean lies exactly in the middle of the distribution. The mean is then the value exactly in the middle of the boundaries of the interval on which the uniform distribution is defined and thus the mean can be determined as the average of the two boundaries.

$$\begin{gathered} \mu =\frac{a+b}{2} \\ =\frac{-1+4}{2} \\ =\frac{3}{2} \\ =1.5 \end{gathered}$$

$$\begin{gathered} P(Y\ge K)=0.25 \\ (4-k)\frac{1}{5}=0.25 \\ \frac{4-k}{5}=0.25 \\ 4-k=1.25 \\ -k=-2.75 \\ k=2.75 \end{gathered}$$