Hi jabeda!
Here are the answers to your questions.
1. In queueing problems, it's usual to assume that the arrival of
customers follows a Poisson distribution. You can find in the
following page several examples of applications of queueing models,
which use a Poisson distribution to model the arrival of customers to
the queue.
M/M/1 Queueing System
http://www.eventhelix.com/RealtimeMantra/CongestionControl/m_m_1_queue.htm
Check also the following page to see the formula for the Poisson
distribution:
The Poisson Distribution
http://stat.tamu.edu/stat30x/notes/node70.html
Thus the probability function is:
P(X=x) = (u^x)*[e^(-u)]/x!
where u is the "arrival rate", and x is the number of arrivals in a
given period. Let's use this in order to answer your question a.
What is the probability that 4 customers will arrive in any given
hour? We know that, on average, there are 2 customer per hour. So the
arrival rate (u) is 2 customers/hour. We want to find the probability
that X=4. Therefore,
P(X=4) = (2^4)*(e^-2)/4! = 16*0.1353/24 = 0.0902
So the probability that 4 customers arrive in any given hour is 0.09.
If we wanted to find the probability of 4 customers arriving in a
2-hour interval, we must first convert the arrival rate (in this case,
since the arrival rate is 2 per hour, than it would be 4 every two
hours) and calculate the probability using this new arrival rate.
Now that you know how to compute this, it's easy to answer part b. The
probability that queueing occurs in any given hour is the probability
that more than 3 customers arrive in any given hour:
P(X>3) = 1-P(X<=3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
Using the formula given above it's easy to compute these
probabilities.
2. Let X be the life of the radio. We know that X is normally
distributed with mean 6 and SD 1.5. In order to find the fraction of
the transistors that last more than 8 years, we have to compute:
P(X >= 8)
Since X is normally distributed, we can use a normal table in order to
find this probability. Normal tables usually give the probability that
a normally distributed random variable with mean 0 and variance 1 (a
*standard* normal distribution) is less than or equal to some number.
Since X doesn't have mean 0 and variance 1, we must first "convert" it
in order to be able to use such table.
In order to convert it, we must substract its mean and divide it by
its SD. Thus,
P(X >= 8)
=P(X-6 >= 8-6)
=P( (X-6)/1.5 >= (8-6)/1.5 )
=P( (X-6)/1.5 >= 1.33... )
Now, (X-6)/1.5 is a normally distributed random variable with mean 0
and variance 1, so it's a standard normally distributed random
variable. Let's call this expression Z, so we must find
P(Z>=1.33...)
We can now use a standard normal table. You con find one in the
following page
Standard Normal Table
http://www.stat.psu.edu/~herbison/stat200/stat200_model_demo/supplements/NormalTable.html
This table gives the probability that Z is less than or equal to some
number. You were asked to find the probability that Z is *greater*
than some number. Therefore, in order to use the table, notice that:
P(Z>=1.33...)
=1 - P(Z<1.33...)
Now we're ready to use the table. Look up the value 1.33 in the
borders of the table (row 1.3, column .03). The probability that
corresponds to this number is 0.9082. So,
P(Z>=1.33...) = 1 - P(Z<1.33...) = 1 - 0.9082 = 0.0918
So the probability that a radio lasts more than 8 years is 0.0918.
In order to answer part b, notice that the manufacturer wants to find
a numberof years such that 20% of the radios last less than that
number. So, if he provides a warranty for that number of years, he
will have to repair 20% of the radios he sells. Mathematically, we
have to find 'a', where 'a' solves:
P( X < a ) = 0.20
Fortunately this can also be done with the standard normal table.
Again, we convert X to standard normal:
P( X < a )
=P( (X-6)/1.5 < (a-6)/1.5)
=P( Z < (a-6)/1.5) = 0.20
So, we now have to look in the "interior" of the table for the number
most close to 0.20 and find what value corresponds to that
probability. In the table given above, the number most close to 0.20
is 0.2005, and the correspondiong z-value is -0.84. Therefore,
(a-6)/1.5 = -0.84
a = 4.74
Thus he must give a warranty of 4.74 years in order to repair 20% of
the radios.
3. In order to answer this question we make use of the Central Limit
Theorem.
Central Limit Theorem
http://davidmlane.com/hyperstat/A14043.html
Basically, the idea is that, for a large sample, the distribution of
the sample mean is similar to the normal distribution, no matter the
distribution of the population where the sample comes from. If the
population from where the sample is taken has mean u and variance s2
(although it needn't have a normal distribution), then the sample mean
(for a large sample) can be approximated with a normal distribution
with mean u and variance s2/n (where n is the sample size).
Let's call M the sample mean of this question. Since it's a large
sample (n is greater than 30, which is the usual limit for a "large"
sample), M will be approximately normally distributed with mean 190
and variance (25*25)/50.
This problem asks you to find:
P(M > 10,000)
Since you know the distribution of M, its mean and its variance, you
can easily compute this probability using the same procedure as in the
previous exercise.
4.a. Again, this question can be easily answered using the procedure
outlined in question 2. You have to find here P(X>80), where X follows
a normal distribution with mean 72 and SD 9.
b. Since this a small sample, you can't use the central limit
theorem for this question. However, it's a property of normal
distributions that no matter the sample size, if the population where
the sample comes from is normal, then the sample mean follows a normal
distribution, again with mean u and variance s2/n. So it's easy again
to compute this probability because you know the distribution of M
(the sample mean), which is normal, with mean 72 and variance
(9*9)/10.
c. If the population weren't normal, we don't know this probability,
because we can't know the distribution of the sample mean. It's not
correct to use the Central Limit Theorem because this is a small
sample. Also, since the population isn't normal, we can't conclude
that the sample mean is normal. Therefore, it's not possible to
compute this probability.
5. The procedure for solving this exercise is exactly the same as the
one for solving the previous ones. Since the population is normal, the
sample mean will also be normal. In all cases, the mean of the sample
mean will be 16.2 (because that's the mean of the population) while
the variance will be 0.12*0.12/1 in part a, 0.12*012/4 in part b and
0.12*0.12/16 in part c.
In all cases, you have to compute
P ( M < 16 )
where M is normally distributed with mean 16.2 and different variance
in each part of the question as described above.
I hope this helps! If you have any doubt regarding my answer, please
don't hesitate to request a clarification. Otherwise, I await your
rating and final comments.
Best wishes!
elmarto |
