Hi mrynot!
In order to answer all the questions here, you'll need a standard
normal distribution table. You can find this in any Statistics book or
online. Here is a link to such table.
Distribution Tables
http://www.statsoft.com/textbook/sttable.html
A standard normal distribution is a normal distribution with mean 0
and variance 1. It's useful because you can easily convert a normally
distributed random variable (r.v.) with any mean and variance into a
standard normally distributed r.v. This conversion works in the
following way. Let X be r.v. that follows a normal distribution with
mean m and standard deviation s. Then
(X - m)/s
is a r.v. that follows the standard normal distribution.
Question 1
Let X be the length of time a listener is tuned to the station. We must find:
a) Prob(X>20)
Using the procedure I explained before, we'll try to leaave the
equation in terms of a standard normally distributed random variable.
Prob( X > 20 )
=Prob( X-15 > 20-15 )
=Prob( (X-15)/3.5 > (20-15)/3.5 )
So that's it, the left-hand side of the inequality is a standard
normally distributed random variable. We'll call it Z. So we must
find:
Prob ( Z > (20-15)/3.5 )
=Prob ( Z > 1.43 )
Since the table I provided in the link gives the probabilites of Z
being smaller than a given number (instead of greater, as in this
equation) and greater than zero, we transform the equation in order to
be able to use this table:
Prob ( Z > 1.43 )
=1 - Prob(Z < 1.43 )
=1 - [ Prob(Z<0) + Prob(0<Z<1.43) ]
Now we can get the value. Prob(Z<0) is 0.5 (because the standard
normal distribution is symmetric around 0) and Prob(0<Z<1.43)=0.4236
(looking up in the table). Therefore,
1 - [ Prob(Z<0) + Prob(0<Z<1.43) ]
=1 - ( 0.5 + 0.4236 )
=0.076
So the probability of a listener staying tuned to the station for more
than 20 minutes is 0.076.
The following exercises are solves in the exact same manner.
b) Prob(X<=20)
=Prob(Z<=(20-15)/3.5)
=0.9236
c) Prob(10<X<12)
= Prob(X<12) - Prob(X<10)
Now you can solve each of these two terms using the previous procedure to get:
Prob(X<12) = 0.195
Prob(X<10) = 0.076
So Prob(10<X<12) = 0.119
Question 2
The procedure is just the same as before
a. Prob(30000<X<35000) = 0.521
So 52.1% of the students have starting salaries between $30,000 and $35,000
b. Prob(X>40000) = 0.004
So 0.4% of the students have starting salaries higher than $40,000
c. Prob(35000<X<40000) = 0.125
So 12.5% of the students have starting slaries between $35,000 and $40,000
Question 3
The random variable involved in this exercise (the number of minor or
major accident during high-speed chases) actually follows a binomial
distribution. The event (accident) has an equal probability of
occuring (0.4) each time the experiment (high-speed chase) is
repeated. We could solve this problem using a binomial distribution
table, but fortunately, the normal disitribution can be used to
approximate the binomial distribution when the sample size is large
enough, and the former is easier to handle.
Normal Approximation to Binomial
http://www.mis.coventry.ac.uk/~nhunt/binomial/normal.html
http://www.ruf.rice.edu/~lane/stat_sim/binom_demo.html
The mean of a binomial disitribution is n*p and its standard deviation
is n*p*(1-p), where n is the number of times the experiment is
repeated (50 in this case) and p is the probability of the event
happening (0.4 in this case). Therefore, the mean of the distribution
is in this case 0.4*50=20 and the standard deviation is 0.4*0.6*50=12.
Finally, since we can use the normal approximation, we proceed just as
usual:
Prob(X>25)
=Prob(Z>(25-20)/12)
=Prob(Z>0.41)
=0.338
Question 4
a. Again, we proceed as before. The claimed amount is 9. So we must find
Prob(X<9)
=Prob(Z<(9-9.2)/0.25)
=Prob(Z<-0.8)
=0.211
b. We repeat the same as before. We must calculate the proportion in
both scenarios.
With m=9.25 and s=0.25:
Prob(X<9)
=Prob(Z<(9-9.25)/0.25)
=0.158
With m=9.2 and s=0.15
Prob(X<9)
=Prob(Z<(9-9.2)/0.15)
=0.091
So reducing the standard deviation is more effective to reduce the
proportion of hams below label weight.
Question 5
This problem appears to be more difficult because it involves a
non-normal distribution, but it's not. We can use the Central Limit
Theorem, which says:
"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 s^2/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 50, as in
this case, is more than enough. The bottom line is that we don't need
the population to be normal in order for the sample mean to follow a
normal distribution.
Let's call this sample mean Xbar. We know, using the theorem, that
Xbar follows a normal distribution with mean 1200 and std. dev.
250/sqrt(50) [sqrt() means square root of], so the std. dev. is 35.35.
With this information, we repeat the same procedure as before:
Prob( Xbar > 950 )
=Prob( Z > (950-1200)/35.35 )
=1
So there is a 100% chance (of course this is an approximation, but its
extremely unlikely to get a sample mean of less than $950) of the
sample mean being greater than $950.
Google search terms
central limit theorem
://www.google.com/search?hl=en&q=central+limit+theorem&btnG=Google+Search
normal distribution table
://www.google.com/search?hl=en&lr=&q=normal+distribution+table&btnG=Search
normal approximation binomial
://www.google.com/search?hl=en&lr=&q=normal+approximation+binomial&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 |
