1. Find P(2 < x < 18).
   [latex]{P}{({2}{<}{x}{<}{18})}={(\text{base})}{(\text{height})}={({18}-{2})}{(\frac{{1}}{{23}})}={(\frac{{16}}{{23}})}.[/latex]
2. Ninety percent of the smiling times fall below the 90th percentile, k, so P(x <k) = 0.90
   [latex]P(x
3. This probability question is a conditional. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know the baby has smiled for more than eight seconds. Find P(x > 12|x > 8) There are two ways to do the problem.
   1. 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 more than eight seconds. Write a new f(x):[latex]{f{{({x})}}}=\frac{{1}}{{{23}-{8}}}=\frac{{1}}{{15}}[/latex]for 8 < x < 23
      
   1. For the second way, use the conditional formula (shown below) with the original distribution X ~ U (0, 23): For this problem, A is (x > 12) and B is (x > 8).
      

1. 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. [latex]{f{{({x})}}}=\frac{{1}}{{{15}-{0}}}=\frac{{1}}{{15}}[/latex] for 0 ≤x ≤ 15. Find P (x < 12.5). Draw a graph.
   [latex]{P}{({x}{<}{k})}={(\text{base})}{(\text{height})}={({12.5}-{0})}{(\frac{{1}}{{15}})}={0.8333}[/latex] The probability a person waits less than 12.5 minutes is 0.8333.
2. [latex]{\mu}=\frac{{{a}+{b}}}{{2}}=\frac{{{15}+{0}}}{{2}}={7.5}[/latex]. On the average, a person must wait 7.5 minutes.[latex]{\sigma}=\sqrt{{\frac{{{({b}-{a})}^{{2}}}}{{12}}}}=\sqrt{{\frac{{{({15}-{0})}^{{2}}}}{{12}}}}={4.3}[/latex]. The standard deviation is 4.3 minutes.
3. Find the 90th percentile. Draw a graph. Let k = the 90th percentile.
   [latex]{P}(xk 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.
   

1. 0.5714
2. This question has a conditional probability. You are asked to find the probability that a nine-year-old eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see previous Example). You must reduce the sample space.
   1. First way: Since you know the child has already been eating the donut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes. Write a new f(x):[latex]{f{{({x})}}}=\frac{{1}}{{{4}-{1.5}}}={25} \text{ for } {1.5}\leq{x}\leq{4}[/latex]. Find P(x > 2|x > 1.5). Draw a graph.
      [latex]{P}{({x}{>}{2}|{x}{>}{1.5})}={(\text{base})}{(\text{new height})}={({4}-{2})}{(\frac{{2}}{{5}})}=\frac{{4}}{{5}}[/latex]
      The probability that a nine-year old child eats a donut in more than two minutes, given that the child has already been eating the donut for more than 1.5 minutes, is [latex]\frac{{4}}{{5}}[/latex].
   2. Second way: Draw the original graph for X ~ U (0.5, 4). Use the conditional formula[latex]{P}{({x}{>}{2}{mid}{x}{>}{1.5})}=\frac{{{P}{({x}{>}{2} \text{ AND } {x}{>}{1.5})}}}{{{P}{({x}{>}{1.5})}}}=\frac{{{P}{({x}{>}{2})}}}{{{P}{({x}{>}{1.5})}}}=\frac{{\frac{{2}}{{3.5}}}}{{\frac{{2.5}}{{3.5}}}}={0.8}=\frac{{4}}{{5}}[/latex]

1. To find f(x): [latex]{f{{({x})}}}=\frac{{1}}{{{4}-{1.5}}}={12.5} \text{ so } {f{{({x})}}}={0.4}[/latex]
   P(x > 2) = (base)(height) = (4 – 2)(0.4) = 0.8
   
   Uniform Distribution between 1.5 and four with shaded area between two and four representing the probability that the repair time
   x is greater than two
2. P(x < 3) = (base)(height) = (3 – 1.5)(0.4) = 0.6. The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between
   x = 1.5 and x = 3. Note that the shaded area starts at x = 1.5 rather than at x = 0; since X ~ U (1.5, 4), x cannot be less than 1.5.
3. 
   Uniform Distribution between 1.5 and four with shaded area between 1.5 and three representing the probability that the repair time x is less than three
4. 
   Uniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% of repair times.
   [latex]P(xk – 1.5)(0.4)0.3 = (k – 1.5) (0.4)[/latex]; Solve to find k: 0.75 = k – 1.5, obtained by dividing both sides by 0.4
   k = 2.25 , obtained by adding 1.5 to both sides. The 30th percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or less.
5. 
   Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times. [latex]P(x>k)[/latex] = 0.25
   [latex]P(x>k)[/latex] = (base)(height) = [latex](4 – k)(0.4)0.25 = (4–k)(0.4)[/latex]; Solve for k: 0.625 = 4 − k, obtained by dividing both sides by 0.4

−3.375 = − k, obtained by subtracting four from both sides: k = 3.375

The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer).

Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 3.375 hours is the 75th percentile of furnace repair times.

• [latex]{\mu}={\frac{a+b}{2}}\text{ and }{\sigma}=\sqrt{\frac{(b-a)^2}{12}}[/latex]
  [latex]{\mu}=\frac{1.5+4}{2}=2.75\text{ hours and }{\sigma}=\sqrt{\frac{(4-1.5)^2}{12}}= 0.7217 \text{ hours}[/latex]