1. Find P(2 < x < 18).
   [latex]{P}{({2}{<}{x}{<}{18})}={(\text{base})}{(\text{height})}={({18}-{2})}{(\frac{{1}}{{23}})}={(\frac{{16}}{{23}})}\approx0.6957[/latex]
2. Ninety percent of the smiling times fall below the 90th percentile, k, so P(x <k) = 0.90
   [latex]P(x

90% of the smile times are at 20.7 or less seconds.

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.

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

[latex]{P}{({x}\geq{12} given x > 8 seconds)}={(\text{base})}{(\text{height})}={({23}-{12})}{(\frac{{1}}{{15}})}={(\frac{{11}}{{15}})}\approx0.7333[/latex]

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}})}\approx{0.8333}[/latex] The probability a person waits less than 12.5 minutes is approximately 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. 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) = 0.6The 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 can not be less than 1.5.
   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
3. 
   Uniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% of repair times.
   P (x < k) = 0.30
   P(x < k) = (base)(height) = (k – 1.5)(0.4)   0.3 = (k – 1.5) (0.4); 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.
4. 
   Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times. P(x > k) = 0.25
   P(x > k) = (base)(height) = (4 – k)(0.4)    0.25 = (4 – k)(0.4); 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]