The Uniform Distribution: example 5.6

a. To find f(x): f (x) = $\frac{1}{4-1.5}$ = $\frac{1}{2.5}$ so f(x) = 0.4

P(x > 2) = (base)(height) = (4 – 2)(0.4) = 0.8

Figure 5.18 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

b. 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. Because X ~ U(1.5, 4), x cannot be less than 1.5.

Figure 5.19 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

c.

Figure 5.20 Uniform distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30 percent 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 30^th percentile of repair times is 2.25 hours. 30 percent of repair times are 2.5 hours or less.

d.

Figure 5.21 Uniform distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25 percent 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 percent of furnace repairs take at least 3.375 hours (3.375 hours or longer).
Note: Since 25 percent of repair times are 3.375 hours or longer, that means that 75 percent of repair times are 3.375 hours or less. 3.375 hours is the 75^th percentile of furnace repair times.

e. $\mu =\frac{a+b}{2}$ and $\sigma =\sqrt{\frac{{(b-a)}^2}{12}}$
$\mu \frac{1.5+4}{2}2.75$ hours and $\sigma =\sqrt{\frac{{(4–1.5)}^2}{12}}=0.7217$ hours