5.7 Continuous random variables: the uniform distribution  (Page 2/4)

0.5714

$\frac{4}{5}$

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

Example 4 figure 1

$P(x< 3)$ = (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 can not be less than 1.5.

Example 4 figure 2

Example 4 figure 3

$P(x< k)=0.30$

$P(x< k)=\left(\text{base}\right)\left(\text{height}\right)=(k-1.5)\cdot \left(0.4\right)$

• 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.

Example 4 figure 4

$P(x> k)=0.25$

$P(x> k)=\left(\text{base}\right)\left(\text{height}\right)=(4-k)\cdot \left(0.4\right)$

• 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 4 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.

$\mu =\frac{a+b}{2}$ and $\sigma =\sqrt{\frac{(b-a{)}^2}{12}}$

$\mu =\frac{1.5+4}{2}=2.75$ hours and $\sigma =\sqrt{\frac{(4-\mathrm{1.5}{)}^2}{12}}=0.7217$ hours