Which of the investment plans in the previous example carries more risk?

To decide which plan carries more risk, we need to look at their variances. Let's begin by calculating the variance separately for each investment plan.

Option A:

x | P(X=x)

-$2000 | 0.2

-$575 | 0.1

$950 | 0.3

$1000 | 0.3

$4000 | 0.1

x ** P(X=x) | x^2 ** P(X=x)

-2000 ** 0.2 = -400 | (-2000)^2 ** 0.2=800,000

-575 ** 0.1 = -57.5 | (-575)^2 ** 0.1 = 33,062.5

905 ** 0.3 = 285 | (950)^2 ** 0.3 = 270,750

1000 ** 0.3 = 300 | (1000)^2 ** 0.3 = 300,000

4000 ** 0.1 = 400 | (4000)^2 ** 0.1 =1,600,000

=527.5 =3,003,812.5

Variance:

standard deviation sign^2 = sum sign [ xi^2 * P(X=xi) ] - Mu^2

=3,003,812.5 - (527.5)^2

=3,003,812.5 - 278,256.25

=2,725,556.25

Standard deviation:

standard deviation sign = square root of "sum sign [ xi^2 * P(X=x) ] - Mu^2"

= square root of 2,725,556.25

=$1650.93

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Option B:

x | P(X=x)

-$1000 | 0.1

-$690 | 0.2

-$100 | 0.2

$1500 | 0.3

$3000 | 0.2

x ** P(X=x) | x^2 ** P(X=x)

-1000 ** 0.1 = -100 | (-1000)^2 ** 0.1 =100,000

-690 ** 0.2 = -138 | (-690)^2 ** 0.2 = 95,220

-100 ** 0.2 = -20 | (-100)^2 ** 0.2 = 2000

1500 ** 0.3 = 450 | (1500)^2 ** 0.3 = 675,000

3000 ** 0.2 = 600 | (3000)^2 ** 0.2 =1,800,000

= 792 =2,672,220

Variance:

standard deviation sign^2 = sum sign [ xi^2 * P(X=xi) ] - Mu^2

=2,672,220 - (792)^2

=2,044,956

Standard deviation:

standard deviation sign = square root of "sum sign [ xi^2 * P(X=x) ] - Mu^2"

=square root of 2,044,956

=$1430.02

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Comparing the standard deviations, since Option A has a standard deviation of $1650.93, and option B has a standard deviation of $1430.02, we can see that not only does Option B have a higher expected value, but its profits vary slightly less than Option A. Therefore, we can conclude that Option B carries a lower amount of risk than Option A. The probability distribution of x successes can be calculated using the classical method of listing the outcomes or simple events of n trials. If a coin is tossed three times, there are eight different possibilities and hence eight simple events. In the following table, the events are grouped according to the number of heads contained in each simple event.

Events | # of heads | Probability

TTT | 0 | 1/8

HTT,THT,TTH| 2 | 3/8

HHT,HTH,THH| 2 | 3/8

HHH | 3 | 1/8

The table produces a probability distribution for the number of heads in 3 tosses of a coin. But this becomes unnecessarily tedious of the value of n is increased. The number of simple events for an experiment with 10 tosses would be 1024, an an experiment with 20 tosses would require listing a staggering 1,048,576.

Fortunately, there is a far simpler method of calculating binomial probabilities:

When all of the binomial distribution conditions are met, the following formula can be used to determine the probability of obtaining x successes out of n trials.

P(X=x) = nCx * P^x (1-p)^(n-x)

where...

x= number of successes

n= number of trials

p= the probability of getting a success on any trial

nCx represents the number of combinations of n objects taken x at a time (without replacement) and is given by...

nCx = n! / x! (n-x)!

where..

n! = n(n-1)(n-2)...1, and 0! =1

ROUNDING RULE!!!!

We will round binomial probabilities to THREE decimal places What is the probability of getting exactly 10 tails in 18 coin tosses?

Solution:

For this problem, there are 18 coin tosses, so n=18. We will say that a success is getting a tail. We want the probability of 10 successes, so x=10. The probability of flipping a tail is 0.5, which means p=0.5. Substituting these values into the binomial formula gives us the following:

In excel:

=BINOM.DIST(x,n,p,cumulative?)

0= not cumulative

1=cumulative

=BINOM.DIST(10,18,0.5,0) = 0.167 Instead of counting the number of occurrences of "successes" in a time interval, there are a number of applications of the Poisson distribution that measure the number of successes in some area or length. The mean number of successes in the area or length will define the parameter, lambda, of the Poisson random variable.

Ex:

The telephone company is considering purchasing optical cable from Optica, Inc. The company wishes to replace approximately 100,000 feet of conventional cable with optical fiber. Since optical fiber is very difficult to repair, it is important that the number of optical cable defects are minimized. Optica claims that the mean number of defects per 200,000 feet of cable is 1. What is the probability that the replaced cable will contain no defects?

Let X= the number of defects in 100,000 feet of optical cable.

Since the rate of defects is given at 1 per 200,000 feet of cable, we will have a rate of 0.5 defects per 100,000 feet of cable. Therefore, lambda= 0.5. Substituting the values into the Poisson distribution function, we have...

P(X=0) = e^lambda ** lambda^x / x! = e^-0.5 **0.5^0 / 0!

= .6065 ;