Introduction for expected value of sample information tutor:
Expected value is the main thought in probability, in an intellect more general than probability itself. The expected value of a real-valued selection variable offers a compute of the center of the distribution of the variable. More considerably, by taking the expected value of various functions of a common random variable, we can calculate a lot of interesting features of its distribution, including spread and correlation. Tutor is a personality working in the education of others, either separately or in group.
Formula for expected value of sample information tutor:
The following formula for expected value of sample information tutor which is used to compute expected
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Solution: Expected value is recognized for the discrete possibility variable by utilize the formula, E(x) = sum xi P (xi) E(x) = 0 (1/13) + 1 (1/13) + 2 (1/13) + 3(1/13) + 4(1/13) E(x) = 0 + 0.0769 + 0.1538 + 0.2307 + 0.3076 E(x) = 0.769 The Expected value is: 0.769
Expected value of sample information tutor - Example 2:
2) Evaluate the expected value to the discrete chance variable 1/6 from 2 to 7.
Solution:
The formula for finding the expected value for a discrete chance variable is E(x) = sum xi P (xi) Here, i = 2 to 7 E(x) = 2 (1/6) + 3 (1/6) + 4(1/6) + 5(1/6) + 6(1/6) + 7(1/6) = 0.3333 + 0.5 + 0.6666 + 0.8333 + 1 + 1.1666 = 4.4998 Answer value is: 4.4998
Additional problems for expected value of sample information tutor:
Expected value of sample information tutor - Example 3:
3) Evaluate the expected value for the discrete possibility variable. (1/18). Where x value begins from 1 to 6.
Solution:
Expected value formula for discrete random variables: E(x) = sum xi P(xi) E(x) = 1 (1/18) + 2 (1/18) + 3 (1/18) + 4(1/18) + 5 (1/18) + 6 (1/18) E(x) = 0.0555 + 0.1111 + 0.1666 + 0.2222 + 0.2777 + 0.3333 E(x) = 1.1664 The Expected value is: 1.1664
Expected value of sample
13) Refer to the table. What is the average number of customers in the queue plus the number being served?
You recently purchased a stock that is expected to earn 12% in a booming economy, 8% in a normal economy and lose 5% in a recessionary economy. There is a 15% probability of a boom, a 75% chance of a normal economy, and a 10% chance of a recession. What is your expected rate of return on this stock?
b) In order to calculate the mean or average for the governors and CEO’s, I added together all the figures and divided that sum by 4 since there
(a) Then mean of the sample and the value of Z with an area of 10% in right tail.
First we find E by doing Zc(standard deviation/square root of number of trials.) Now we add and subtract that number from the mean income to find both endpoints. The Zc of 95% is 1.96 so we would do
E[RA ] = 0.3 × 0.07 + 0.4 × 0.06 + 0.3 × (−0.08) = 0.021 = 2.1%,
b. What would Mrs. Beach have to deposit if she were to use common stock and earned an average rate of return of 11%.
distribution with an average of 2 per 30 second period. What is the probability of having
(60% are 30 s) + (15% are 40 s) + (5% are 240 s) + (3% are 50 s)
28. The Polo Development Firm is building a shopping center. It has informed renters that their rental spaces will be ready for occupancy in 19 months. If the expected time until the shopping center is completed is estimated to be 14 months, with a standard deviation of 4 months, what is the probability that the renters will not be able to occupy in 19 months?
1. What is the expected customer lifetime value of a newly acquired customer? Use an annual discount rate of 10%.
2.56 0.60 48.86 3.64 73.43 40.16 63.63 102.47 4.06 1.37 0.63 0.56 4.63 1.25 34.33% 11.19% 5.70% 9.98% 21.17%
P = $40({1 – [1/(1 + .0475)]26 } / .0475) + $1,000[1 / (1 + .0475)26]
Assume that under no unusual circumstances (no storm), Jaeger sells 1,000 cases of Riesling. Consider different cases: 1. Jaeger harvests grapes in anticipations of storm. Then the total revenue will be equal to 12×1000×$2.85 = $34200. 2. Jaeger doesn’t harvest and there is no storm with 50% chance. 2.1. With 40% chance, sugar concentration is 25%, then the total revenue is 12 × 1000 × $3.50 = $42000 2.2. With 40% chance, sugar concentration is 20%, then the total revenue is 12 × 1000 × $3.00 = $36000 2.3. With 20% chance, sugar concentration is below 20%, then the total revenue is 12 × 1000 × $2.50 = $30000 3. There is storm with 50% chance 3.1. Storm causes botrytis
By estimating the average number of students who will visit the tutor room during the opening hours, we can determine whether the Department of Finance and Actuarial Science can close down the tutor room during some specific hours so that they can reduce the hiring expenses. Firstly, we need to identify the distribution needed for this test. 3As the numbers of students who will visit the tutor room during the opening hours are subject to the timing students usually study and all this and would not affect one another, this random variable is independently and