i) Find the probability that a randomly selected furnace repair requires more than 2 hours. ii) Find the probability that a randomly selected furnace repair requires less than 3 hours. iii) Find the mean and standard deviation.

i) Probability that a randomly selected furnace repair require more than 2 hours

Let x be random variable that denotes the amount of time a repairman needs to fix a furnace. When X is uniformly distributed between (1.5, 4 ) (a, b)

So, required probablity is - $1 - P(0 \lt X \lt 2)$

$\begin{aligned} \text{We know, } P(0 \lt X \lt 2) &= F(2) - F(0) \\ &= \frac{2-0}{b-a} \\ &= \frac{2}{4-1.5} \\ &= \frac{2}{2.5} \\ &= 0.8 \end{aligned}$

$\therefore$ Probability that a randomly selected furnace repair requires more than two hours = $1 - 0.8 = 0.2$

ii) Find probability that randomly selected furnace repair requires less than 3 hours

$\begin{aligned} &=P(\alpha x \lt 3) \\ &=F(3)-F(0) \\ &=\frac{3-0}{b-a} \\ &=\frac{3}{4-1 \cdot 5} \\ &=\frac{3}{2 \cdot 5} \\ &=1 \cdot 2 \end{aligned}$

iii) Find the mean and standard deviation

mean, $E(x)=\frac{a+b}{2}=\frac{1 \cdot 5+4}{2}=\frac{45}{2} = 2.25$

$\begin{aligned} \text { variance, } V(x) &=\frac{(b-a)^{2}}{12} \\ &=\frac{(4-1.5)^{2}}{12}=\frac{(2.5)^{2}}{12} \\ &=\frac{6.25}{12}=0.520 \end{aligned}$

$\begin{aligned} \therefore \text { Standard Deviation } &=\sqrt{\text{Variance}} \\ &=\sqrt{0.520} \\ &=0.721 \end{aligned}$