Hello!
Here are the answers to your questions.
Question 1
I'll assume here that the samples are taken "without replacement".
That is, after taking the first of the four sample items, we don't
replace it to the population, so that we now have 14 items to choose
from.
In these cases, the appropiate distribution function that should be
used is the hypergeometric distribution. This distribution is good for
describing problems that "...ask for the probability that i out of N
balls drawn are "good" from an urn that contains n "good" balls and m
"bad" balls"
MathWorld -- Hypergeometric Distribution
http://mathworld.wolfram.com/HypergeometricDistribution.html
In this link, you find the formula for the probability you're looking
for. Using the notation described there, we have that, for your
problem:
n = 10 ("good" items)
m = 5 ("bad" items)
N = 4 (sample size)
i = 3 (you're interested in the prob. of drawing 3 "good" items)
In order to find the probability, just plug these values into the
formula shown in the previous link. This will give you that the
probability of drawing 3 good items is 0.4395.
Question 2
The appropiate distribution to use in this case is the Poisson
distribution. This discrete distribution is often used to model events
that can happen with the same probability at any point in time. Here,
we'll stretch this interpretation and think "time" as being each
chapter. Thus, in this experiment, the publishers might make mistakes
any numer of times, at any point, while writing a chapter. We know,
however, that the "rate" of errors is 0.8 errors/chapter.
Poisson distribution -- Wikipedia
http://en.wikipedia.org/wiki/Poisson_distribution
As you can see in the link, tha only parameter of the Poisson
distribution is lambda, the rate at which the events happens. In this
case, lambda=0.8.
We want to find here the probability that there are less than two
errors in a given chapter. So, calling X to the number of errors, we
want to find P(X<2). But,
P(X<2) = P(X=0) + P(X=1)
(the number of errors can't be less than zero)
So, we use the formula shoen in the link above, plugging lambda=0.8,
and k=0 for P(X=0), and k=1 for P(X=1) to obtain:
P(X<2) = P(X=0) + P(X=1)
= 0.4493 + 0.3594
= 0.8087
SO the probability of making less than 2 mistakes is 0.8087.
Question 3
Calling X to the starting salary for college grads, we have that
X ~ N(31280,3000)
(that is, the salary follows a normal distribution with mean 31280 and
std. dev. 3000)
a. We want to find here
P(30000 < X < 35000)
=P(X < 35000) - P(X < 30000)
In order to find this probability, we need to use a standard normal
distribution table. You can find one in any Statistics book or at the
following link
z-distribution
http://www.math2.org/math/stat/distributions/z-dist.htm
Since the table gives the probabilities for the standard normal
distribution, we must first "convert" X to a standard normal
distribution by subtracting its mean and dividing it by its standard
deviation. So we have that:
P( X < 35000 )
=P( (X-31280)/3000 < (35000-31280)/3000 )
=P( Z < 1.24 )
I renamed (X-31280)/3000 as Z because the variable "(X-61)/9" follows
a standard normal distribution (that is, a normal distribution with
mean 0 and variance 1).
Now we look up the value 1.24 in the table, getting that:
P(Z < 1.24) = 0.8925
Therefore, P(X<35000)=0.8925.
Now we need to find P(X<30000). In order to do it, we follow exactly
the same procedure as before. We get that P(X<30000)=0.3348.
Therefore,
P(30000 < X < 35000)
=P(X < 35000) - P(X < 30000)
=0.8925 - 0.3348
=0.5577
Thus 55.77% of college grads gat starting salaries between $30,000 and $35,000.
b. We want to find P(X>40000). We follow the same procedure as before
(i.e., "converting" X to a standard normal distribution and looking up
in the table. Be careful that in this case you will be looking at Z
being greater than some number, rather than Z being smaller than a
number, like in point a)
We get here that P(X>40000)=0.0018. So 0.18% of college grads get
starting salaries greater than $40,000
c. Here we want to find P(35000 < X < 40000). Again, we follow the
same procedure. We find here that 10.56% of college grads get starting
salaries between $35,000 and $40,000
[If you are having trouble with points b and c, please don't hesitate
to request clarification]
Question 4
The appropiate distribution to use in this question is the binomial
distribution. I'll assume here that the fact that 80% of the rooms are
occupied during september implies that the probability of each
individual room being occupied is 80%, or 0.8. Therefore, given that
there are 800 rooms, we know that the number of occupied rooms during
september follows a binomial distribution with parameters n=800 and
p=0.8.
Binomial Distribution
http://www.stat.tamu.edu/stat30x/notes/node66.html
Calling X to the number of occupied rooms, we want to find:
P(X>=665)
Since there is no easy commulative distribution function for the
binomial distribution, we would need to calculate a bunch of values in
order to get this probability (
P(X=665)+P(X=666)+P(X=667)+...+P(X=800) ). Fortunately, when n is
large, we can approximate the binomial distribution with the normal
distribution. So we first need to find the mean and std. dev. of X.
Since the mean of the binomial distribution is n*p, then the mean of X
is 0.8*800=640. Since the std. dev. of the binomial distribution is
sqrt(n*p*(1-p)), then the std. dev. of X is sqrt(800*0.8*0.2)=11.31.
Therefore, the distribution of X can be approximated with a normal
distribution with mean 640 and std. dev. 11.31. So now we follow the
same procedure as in the previous question:
P(X >= 665)
=P( (X-640)/11.31 >= (665-640)/11.31 )
=P( Z >= 2.21 )
We again use the table to find that this probability is 0.0135. So
that's the probability that 665 or more rooms are occupied.
Question 5
In order to answer this question, we'll need to make use of the
Central Limit Theorem:
"The central limit theorem states that given a distribution with a
mean m and variance s2, the sampling distribution of the mean
approaches a normal distribution with a mean (m) and a variance s2/N
as N, the sample size, increases."
Central Limit Theorem
http://davidmlane.com/hyperstat/A14043.html
Sample sizes larger than 30 can usually be considered large enough for
the sample mean to have an almost normal distribution. So the sample
size of 40 in this question is more than enough to use this theorem.
This theorem is very powerful. It says that, no matter what the
distribution of X, the distribution of the sample mean of X will be
normal with mean m and std.dev s/sqrt(N). So, in your case, the sample
mean will follow a approximately a normal distribution, with mean 330.
We'll see about the standard deviation later.
A. As we've said, no assumptions need to be made about the shape of
the population. With a large enough sample, the sample mean will
follow a normal distribution.
B. The standard error of the mean can be calculated as the standard
error of the population divided by the square root of the sample size.
In this case, however, we'll use the sample std. dev., since we don't
know the one of the population. So the standard error of the mean is
80/sqrt(40)=12.64.
C. In order to answer this and the following questions, we follow the
same procedure as in previous questions. Let's call Xbar to the sample
mean.
P(Xbar > 320)
=P( (Xbar-330)/12.64 > (320-330)/12.64)
=P( Z > -0.79 )
Notice now that in the left-hand side of the inequality, we are
dividing by the sample std. dev. rather than the population std. dev.
This means that, being strict, we should use the T distribution
instead of the standard normal distribution Z. However, when the
sample size is large, the T distribution is very similar to the
standard normal distribution. So, for simplicity, I'll just compute
the results using the standard normal distribution.
So, checking -0.79 in the table, we get that:
P(Z > -0.79) = 0.785
The probability that the sample mean is greater than 320 minutes is 0.785.
D. We follow the same procedure. We get that this probability is 0.728.
E. Again, the same procedure. The probability is 0.0567.
Google search terms
hypergeometric distribution
://www.google.com/search?hl=en&q=hypergeometric+distribution&spell=1
poisson distribution
://www.google.com/search?hl=en&lr=&q=poisson+distribution
binomial distribution
://www.google.com/search?hl=en&lr=&q=binomial+distribution
normal distribution table
://www.google.com/search?hl=en&lr=&q=normal+distribution+table&btnG=Search
"central limit" theorem
://www.google.com/search?hl=en&lr=&q=%22central+limit%22+theorem&btnG=Search
I hope this helps! If you have any questions regarding my answer,
please don't hesitate to request a clarification. Otherwise I await
your rating and final comments.
Best wishes!
elmarto |
