Question

In: Statistics and Probability

Please find the Standard Deviation of each option Option A Option B Payout Probability Payout Probability...

Please find the Standard Deviation of each option

Option A

Option B

Payout

Probability

Payout

Probability

$200

0.05

$100

0.01

$ 50

0.10

$ 55

0.14

$ 25

0.15

$ 35

0.15

$    5

0.20

$ 29

0.20

$    1

0.50

$    1

0.50

Solutions

Expert Solution

MEAN:

Let x be a discrete r.v. with values i=1,2,...,n represent the probability distribution of a r.v. x denote by E(X)  is defined as

VARIANCE:

Let x be a discrete r.v. with if i=1,2,...,n represent the probability distribution of a r.v. x denote by is defined as,

STANDARD DEVIATION:

The positive square root of the variance is called the standard deviation (SD).

OPTION A:

x pi pi*x pi*x*x
200 0.05 10 2000
50 0.1 5 250
25 0.15 3.75 93.75
5 0.2 1 5
1 0.5 0.5 0.5
total 1 20.25 2349.25
E(X^2)= 2349.25
Mean=E(X)= 20.25
(E(X))^2= 410.0625
Variance=V(X)= 1939.188
Standard Deviation= 44.03621

Standard deviation of option A is 44.03621

OPTION B:

x pi pi*x pi*x*x
100 0.01 1 100
55 0.14 7.7 423.5
35 0.15 5.25 183.75
29 0.2 5.8 168.2
1 0.5 0.5 0.5
Total 1 20.25 875.95
E(X^2)= 875.95
Mean=E(X)= 20.25
(E(X))^2= 410.0625
Variance=V(X)= 465.8875
Standard Deviation= 21.58443

Standard deviation of option B is 21.58443


Related Solutions

1. Find the mean and standard deviation for option 1 and option 2 Option 1: Year...
1. Find the mean and standard deviation for option 1 and option 2 Option 1: Year Return 1 15 % 2 -12 % 3 8 % 4 11 % Option 2: Prob. Return 0.05 16 % 0.40 8 % 0.25 -6 % 0.30 7 %
Find the missing value required to create a probability distribution, then find the standard deviation for...
Find the missing value required to create a probability distribution, then find the standard deviation for the given probability distribution. Round to the nearest hundredth. x / P(x) //////// 0 / 0.2 1 / 2 / 0.13 3 / 0.03 4 / 0.05
In a normal distribution with mean = 27 and standard deviation = 4 Find the probability...
In a normal distribution with mean = 27 and standard deviation = 4 Find the probability for a.) 23 < x < 31 b.) 27<x<35 c.) 25 < x < 30 d.) x>26 e.) x < 24
Find the Standard Score (z-score) round to 2 decimal places and the standard deviation Option Amazon...
Find the Standard Score (z-score) round to 2 decimal places and the standard deviation Option Amazon Description of Item Price Standard Score (z-score) (round to 2 decimal places) 1 Beginner Acoustic Guitar Set $84.99 2 Brazilian Double Hammock $27.99 3 MMIZOO Resistance Loop Exercise Bands (set of 5) $9.99 4 Big and Tall Office Chair $185.99 5 Adjustable Laptop Bedstand Table $32.98 6 HP Envy Pro 6455 Wireless All-in-One Printer $149.99 7 Apple Watch Series 3 $169.00 8 Waterproof Portable...
Find the probability that the Normal random variable with mean 20 and standard deviation 3.2 will...
Find the probability that the Normal random variable with mean 20 and standard deviation 3.2 will generate an outlier (outside the inner fences) observation. Remember that the lower (upper) inner fence is 1.5*IQR below (above) the first (third) quartile. a. 0.0035          b. 0.0051          c. 0.0058          d. 0.0062          e. 0.0070
Find the SS, the variance, and the Standard Deviation of Sample A Please be sure to...
Find the SS, the variance, and the Standard Deviation of Sample A Please be sure to show your entire work. For each step. Sample A 8 10 18 1 6 11 9 10 10 7
Using the standard normal distribution, find each probability. a) P(0 < z < 2.23) b) P...
Using the standard normal distribution, find each probability. a) P(0 < z < 2.23) b) P (-1.75 < z < 0) c) P (-1.48 < z < 1.68) d) P (1.22 < z < 1.77) e) P (-2.31 < z < 0.32)
a. The standard deviation of a sample of data was 14. Find the variance. B. Given...
a. The standard deviation of a sample of data was 14. Find the variance. B. Given the sample data 5, 0, 7, 5, 9, and 6, find its range C. A new​ weight-loss program claims that participants will lose an average of more than 10 pounds after completing it. The data table shows the weights of five individuals before and after the program. 1 2 3 4 5 weight before 264 220 285 264 195 weight after 240 223 267...
The population mean and standard deviation are given below. Find the required probability and determine whether...
The population mean and standard deviation are given below. Find the required probability and determine whether the given sample mean would be considered unusual. For a sample of n=63 , find the probability of a sample mean being less than 19.8 if μ=20 and σ=1.33. For a sample of n=63 , the probability of a sample mean being less than 19.8 if μ=20 and σ=1.33 is ______. (Round to four decimal places as needed.) Would the given sample mean be...
The population mean and standard deviation are given below. Find the indicated probability and determine whether...
The population mean and standard deviation are given below. Find the indicated probability and determine whether a sample mean in the given range below would be considered unusual. If​ convenient, use technology to find the probability. For a sample of nequals=3737​, find the probability of a sample mean being less than 12 comma 75212,752 or greater than 12 comma 75512,755 when muμequals=12 comma 75212,752 and sigmaσequals=1.31.3. For the given​ sample, the probability of a sample mean being less than 12...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT