Question

(a) The time X minutes , taken by Fred Fast to install a satellite dish may be assumed to be a random variable with mean 134 and standard deviation 16.

i) Determine P(X<150) (3Marks)
ii) Determine to one dp the time exceeded by 10% of installations. (4Marks)

b) The time Y minutes taken by Sid Slow to install a satellite dish may also be assumed a normal random variable but with

P(Y<170)=0.14 and P(Y>200)=0.03

Determine to the nearest minute value for the mean and standard deviation of Y. (6Marks)

Please show all working thanks

Answer 1

Solution

a).

i). P(X <150)

$$= \phi({x- \mu \over \sigma}) = \phi ({150-134 \over 16})= \phi (1)$$

$=0.8413$

ii). P(X >x) =0.1

$$P(X>x) =1-P(X \le x) =0.1$$ $$P(X \le x) =0.9$$ $$\phi ({x-134 \over 16})=0.9$$

$${x-134 \over 16}=1.28$$ $$x=1.28(16)+134$$

$=154.5 \space minutes$

b). P (Y<170)=0.14 and P(Y>200)=0.03

$$\phi ({170 - \mu \over \sigma}) =0.14$$

$${170 - \mu \over \sigma} =-1.08$$ $$170- \mu = - 1.08 \sigma \space........ (i)$$

$$P(Y>200)=1-P(Y \le 200)=0.03$$

$$\phi ({200 - \mu \over \sigma}) =1-0.03=0.97$$

$${200 - \mu \over \sigma} =1.88$$ $$200-\mu =1.88 \sigma \space...... (ii)$$

Solve eq(i) and (ii) simultaneously.

Subtracting eq(i) from (ii) yields :

$$30=2.96 \sigma$$ $$\therefore \sigma \approx10 \space minutes$$ $$\mu =200-1.88(10)$$ $$\therefore \mu \approx 181 \space minutes$$