6.2 The uniform distribution

Find $P(2< x< 18)$ .

$P(2< x< 18)=\left(\text{base}\right)\left(\text{height}\right)=(18-2)\cdot \frac{1}{23}=\frac{16}{23}$ .

Ninety percent of the smiling times fall below the 90th percentile, $k$ , so $P(x< k)=0.90$

$P(x< k)=0.90$

$\left(\text{base}\right)\left(\text{height}\right)=0.90$

$(k-0)\cdot \frac{1}{23}=0.90$

$k=23\cdot 0.90=20.7$

Find $P(x> 12|x> 8)$ There are two ways to do the problem. For the first way , use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled morethan 8 seconds.

Write a new $f(x)$ : $f(x)=\frac{1}{23-8}=\frac{1}{15}$

for $8< x< 23$

$P(x> 12|x> 8)=(23-12)\cdot \frac{1}{15}=\frac{11}{15}$

For the second way, use the conditional formula from Probability Topics with the original distribution $X$ ~ $U(0,23)$ :

$P\left(A\right|B)=\frac{P\left(A\text{AND}B\right)}{P\left(B\right)}$ For this problem, $A$ is $(x> 12)$ and $B$ is $(x> 8)$ .

So, $P(x> 12|x> 8)=\frac{(x>12\text{AND}x>8)}{P(x> 8)}=\frac{P(x> 12)}{P(x> 8)}=\frac{\frac{11}{23}}{\frac{15}{23}}=0.733$

Let $X$ = the number of minutes a person must wait for a bus. $a$ = 0 and $b$ = 15. $x~U(0,15)$ . Write the probability density function. $f(x)=\frac{1}{15-0}=\frac{1}{15}$ for $0\le x\le 15$ .

Find $P(x< 12.5)$ . Draw a graph.

$P(x< k)=\left(\text{base}\right)\left(\text{height}\right)=(12.5-0)\cdot \frac{1}{15}=0.8333$

The probability a person waits less than 12.5 minutes is 0.8333.

$\mu =\frac{a+b}{2}=\frac{15+0}{2}=7.5$ . On the average, a person must wait 7.5 minutes.

$\sigma =\sqrt{\frac{(b-a{)}^2}{12}}=\sqrt{\frac{(\mathrm{15}-0{)}^2}{12}}=4.3$ . The Standard deviation is 4.3 minutes.

Find the 90th percentile. Draw a graph. Let $k$ = the 90th percentile.

$P(x< k)=\left(\text{base}\right)\left(\text{height}\right)=(k-0)\cdot \left(\frac{1}{15}\right)$

$0.90=k\cdot \frac{1}{15}$

$\mathrm{k\; =\; (0.90)(15)\; =\; 13.5}$

$k$ is sometimes called a critical value.

The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes.