Download Probability distributions with solutions

Statistical Distributions
Example 1:
Solution:
Example 3:
Solution:
Example 4:
Solution:
, so therefore
Expected Value of a discrete random variable X
Expected value is the same as finding the mean. We can do this by using a modification of the mean
equation,
which looks like this:
Example 1:
Solution using the equation:
We can also use technology to solve this:
Ti-84 – Put x under L1 and P(X=x) under L2 under STAT. After this is entered go to STAT->CALC and select
1-Var Stats then enter L1, L2 after 1-Var Stats then ENTER.
Example 2:
Solution:
1
0.1
2
0.3
3
0.6
(You can use your calculator to find the standard deviation).
Example 3:
Solution:
1
1/36
2
3/36
3
5/36
4
7/36
5
9/36
6
11/36
Example 4:
Solution:
a)
b)
c) Since you spend $2 on a ticket and expect to win $1.7 you actually lose $0.3 per ticket on average.
Binomial Distribution
We often write a binomial distribution as
If we go back to Example 1 we can use this method since this is a binomial distribution. The number of
trials is 3 (n = 3) and the probability of success is getting a six (p = 1/6). So if we want to find the
probability of getting 2 sixes we can use the following setup:
We can also use the Ti-84 calculator going under 2nd VARS (DISTR)->binompdf . It is set up like this
binompdf(n,p,r) so for our question it will be binompdf(3,1/6,2) = 0.0694.
On the nspire go to menu->Probability->Distributions->Binomial Pdf
Note as well that Binomial Cdf represents the cumulative probability for
. So if
then
.
On the calculator we would enter Ti-84: binomcdf(3,1/6,2) or in Nspire it would be binomCdf(3,1/6,0,2).
Example 1:
a) P(X=6)=binompdf(6,0.5,6)=0.0156
b) P(X=2)=binompdf(6,0.5,2)=0.234
c) P(X
d) P(X
)=binomcdf(6,0.5,1)=0.109
)=1 - P(X
=0.344
)=1-binomcdf(6,0.5,3)
Example 2:
a) P(X=7)=binompdf(7,0.6,7)=0.0280
b)
c) P(X=6)=binompdf(7,0.6,6)=0.131
d) P(X
)=1 - P(X
=0.710
)=1-binomcdf(7,0.6,3)
For binomial distributions where n is the number of trials and p is the probability of success:
Mean of x is E(X)=
Variance of x is Var(X)=
and therefore the standard deviation is
Example 8:
a)
0
0.0156
b)
c)
1
0.0938
2
0.2344
3
0.3125
4
0.2344
5
0.0938
6
0.0156
Example 9:
Solution:
If the random variable X is normally distributed it can be written as
.
In this diagram
there are 3 normal
distributions with
the same standard
deviation but not
the same mean.
In this diagram
there are 3 normal
distributions with
the same mean but
not the same
standard deviation.
Example 4:
a)
b)
c)
= 0.3413 + 0.1359 =0.4772
= 0.3413 + 0.1359 + 0.0215 =0.4987
= 0.9544 + 0.0215 =0.9759
In statistics we use a standard normal distribution where the mean is 0 and a standard deviation is 1.
We use Z as the random variable so
under a standard normal curve
. We can use technology to find the area (probability)
.
Using your GDC:
Ti-84: 2nd VARS -> normalcdf(Lower Bound, Upper Bound)
Nspire: menu->Statistics(6)->Distributions(5)->Normal Cdf (Lower Bound, Upper Bound)
Example 1: Given
a)
b)
find the following:
c)
d)
Solutions:
a) 0.5
b) 0.159 or 1 – (0.5 + 0.3413) c) 0.819
d) 0.961
Since most distributions do not match the shape of the standard normal distribution we can
convert it using the equation:
, if
)
To solve most of these types of questions do the following: (IB often requires this as work)
1. Draw a graph of the distribution.
2. Convert x to z using the equation.
You can also use your GDC:
Ti-84: 2nd VARS -> normalcdf(Lower Bound, Upper Bound, mean, s.d.)
Nspire: menu->Statistics(6)->Distributions(5)->Normal Cdf (Lower Bound, Upper Bound,
mean, s.d)
So if you wanted to find the probability that X was between 1.3 and 4.5 if
) then
you could enter in your calculator normalcdf(1.3,4.5,1.4,2.24) or 0.435. (note that you
should only use this if you do not need to show work)
Example 2:
If Z has a standard normal distribution find a:
a)
b)
Solutions: (make a sketch for each)
a) invNorm(0.881)=1.18
b) 1 – 0.92 = 0.08, therefore invNorm(0.08)=-1.41
Example 3:
Example 4:
Solution:
Example 5:
Solution:
Example 6:
Solutions:
a)
Therefore using the standard normal equations we get:
and using systems of equations we can find that:
b) The probability that the token will not work is p = 0.05. So this is a binomial probability:
and we want to find
, which is 0.7358.