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The previous problem is an example of the uniform probability distribution .
Illustrate the uniform distribution . The data that follows are 55 smiling times, in seconds, of an eight-week old baby.
10.4 | 19.6 | 18.8 | 13.9 | 17.8 | 16.8 | 21.6 | 17.9 | 12.5 | 11.1 | 4.9 |
12.8 | 14.8 | 22.8 | 20.0 | 15.9 | 16.3 | 13.4 | 17.1 | 14.5 | 19.0 | 22.8 |
1.3 | 0.7 | 8.9 | 11.9 | 10.9 | 7.3 | 5.9 | 3.7 | 17.9 | 19.2 | 9.8 |
5.8 | 6.9 | 2.6 | 5.8 | 21.7 | 11.8 | 3.4 | 2.1 | 4.5 | 6.3 | 10.7 |
8.9 | 9.4 | 9.4 | 7.6 | 10.0 | 3.3 | 6.7 | 7.8 | 11.6 | 13.8 | 18.6 |
sample mean = 11.49 and sample standard deviation = 6.23
We will assume that the smiling times, in seconds, follow a uniform distribution between 0 and 23 seconds, inclusive. This means that any smiling time from 0 to and including 23 secondsis equally likely . The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.
Let $X$ = length, in seconds, of an eight-week old baby's smile.
The notation for the uniform distribution is
$X$ ~ $U\left(\mathrm{a,}b\right)$ where $a$ = the lowest value of $x$ and $b$ = the highest value of $x$ .
The probability density function is $f(x)=\frac{1}{b-a}$ for $a\le x\le b$ .
For this example, $x$ ~ $U\left(\mathrm{0,}23\right)$ and $f(x)=\frac{1}{23-0}$ for $0\le x\le 23$ .
Formulas for the theoretical mean and standard deviation are
$\mu =\frac{a+b}{2}$ and $\sigma =\sqrt{\frac{(b-a{)}^{2}}{12}}$
For this problem, the theoretical mean and standard deviation are
$\mu =\frac{0+23}{2}=11.50$ seconds and $\sigma =\sqrt{\frac{(\mathrm{23}-0{)}^{2}}{12}}=6.64$ seconds
Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation.
What is the probability that a randomly chosen eight-week old baby smiles between 2 and 18 seconds?
Find $P(2< x< 18)$ .
$P(2< x< 18)=\left(\text{base}\right)\left(\text{height}\right)=(18-2)\cdot \frac{1}{23}=\frac{16}{23}$ .
Find the 90th percentile for an eight week old baby's smiling time.
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 the probability that a random eight week old baby smiles more than 12 seconds KNOWING that the baby smiles MORE THAN 8 SECONDS .
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 $X$ ~ $U(0,23)$ :
$P\left(A\right|B)=\frac{P\left(A\phantom{\rule{2pt}{0ex}}\text{AND}\phantom{\rule{2pt}{0ex}}B\right)}{P\left(B\right)}$ For this problem, $A$ is $(x> 12)$ and $B$ is $(x> 8)$ .
So, $P(x> 12|x> 8)=\frac{(x>12\phantom{\rule{2pt}{0ex}}\text{AND}\phantom{\rule{2pt}{0ex}}x>8)}{P(x> 8)}=\frac{P(x> 12)}{P(x> 8)}=\frac{\frac{11}{23}}{\frac{15}{23}}=0.733$
Uniform : The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between 0 and 15 minutes, inclusive.
What is the probability that a person waits fewer than 12.5 minutes?
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.
On the average, how long must a person wait?
Find the mean, $\mu $ , and the standard deviation, $\sigma $ .
$\mu =\frac{a+b}{2}=\frac{15+0}{2}=7.5$ . 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.
Ninety percent of the time, the time a person must wait falls below what value?
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.
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