Probability Question

[WeeG]

Hello,

I need a wee help with this question...

"A company that repair computers go with the next sale:
If the waiting time of a customer for a technician is at least as long as the time
of the repair itself, the customer will pay 55$. On the other hand, if the waiting
time for a technician is less that the time of the repair, the customer will pay
95$.
The waiting time for a technician is a normal random variable with mean of 8
hours and standard deviation of 75 minutes.
The time of repair is also a normal random variable with mean of 6 hours and
a standard deviation of 150 minutes.
Assume that the times are independent."

1. Calculate the probability that a customer will pay 55$.
2. Calculate the variance of payment of a customer to the company.
3. The company has decided to give bonus to it's workers if the profit from
repairing computers will be more than 7500$. How many customers do they
need to serve so the probability of getting the bonus will be more than
0.67 ?

I wouldn't know where to start, I mean, I think I have some clue on "1", but not on "2" and "3" ... help !! :o

[BGM]

Some hints:

1. You can evaluate the probability. Note that the sum/difference of
independent normal random variable is still normally distributed.

2. Let the answer in part 1 be p. Note the payment of a customer
is actually a "Bernoulli" (2 support point: 55 and 95) random variable.
You can evaluate the variance using formula easily.

3. The sum of i.i.d. Bernoulli random variable is Binomially distributed.
Here you can approximate the probability with normal approximation as well.

4. To evaluate the normal probabilities you may need to well transformed the variable into a standard one.

[WeeG]

Thank you ! Can you please give me a further assistance in part 3 ?

What I did so far is:

1. (X-Y)~N(2,8.5)
P(X>=Y)=P(X-Y>=0)=1-P(X-Y<0)=1-P(Z<-0.9285)=1-0.246672=0.753328

2. T-payment:
P(T=55)=0.7533 P(T=95)=0.2466
E(T)=64.8585 E(T^2)=4505.2975 Var(T)=297.6725 S.D(T)=17.2531

3. y-Sum
I need to find n for: P(Y>=7500)>0.67
I am stuck....
:confused:

[BGM]

Since Y = \sum_{i=1}^n T_i is an i.i.d. sum,
you may try to use normal approximation to evaluate the probability

\Pr\{Y \geq 7500 \}

= \Pr\left\{\frac {Y - E[Y]} {SD[Y]} \geq \frac {7500 - E[Y]} {SD[Y]} \right\}

\approx \Pr\left\{Z \geq \frac {7500 - E[Y]} {SD[Y]} \right\}

Note both E[Y] and SD[Y] is in terms of n
and this approximation holds well only when n is large.
You can look for the normal quantile and try to solve for the n.


In case your solution is small, you may try to find the exact one.

Note T_i = (95 - 55)B_i + 55 = 40B_i + 55
where B_i \sim Bernoulli(p)

\Rightarrow Y = \sum_{i=1}^n T_i
= \sum_{i=1}^n (40B_i + 55)
= 40 \sum_{i=1}^n B_i + 55n

Note W = \sum_{i=1}^n B_i \sim Binomial(n, p)

So now we have
\Pr\{Y \geq 7500\} = \Pr\{40W + 55n \geq 7500 \}
= \Pr\left\{W \geq \frac {375} {2} - \frac {11n} {8} \right\} > 0.67

As your n is small, you can numerically search for the solution with your
computer, starting n = 1, 2, 3, ...

Using your answers in part 1 and 2
with method 1, n is 117
and with method 2, n is 90.

See if I make any careless mistake :)

[WeeG]

Sorry to bother you about this one :o

I think your first method is the one, because I can't use a computer...

Would you mind explain how you got 117, I tried calculating and didn't get the correct result...

Thank you

[BGM]

According to your answers,

p = 0.753328, E[Y] = 64.8585n, SD[Y] = 17.2531\sqrt{n}

Note \Pr\{Z \geq z\} > 0.67 \iff z < -0.439913165673234

So we have \frac {7500 - 64.8585n} {17.2531\sqrt{n}} <
-0.439913165673234

\Rightarrow 64.8585(\sqrt{n})^2 - 7.58986583867687\sqrt{n} - 7500
> 0


\Rightarrow \sqrt{n} > 10.8121043263598 ~
\rm{or} ~ \sqrt{n} < -10.6950824119049 ~ \rm{(rejected)}

\Rightarrow n > 116.901599964089

[Link]

I'd check #1 again. The distribution of a difference of two independent distributions should have a SD of sqrt(SD1^2 + SD2^2).

[BGM]

Actually I just trust WeeG answer provided in the first 2 parts without checking them.

If there is any change in the answers, you can just do the last part again easily :)