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Aneli
4 days ago
6

Jackson goes to the gym 0, 2, or 3 days per week, depending on work demands. The expected value of the number of days per week t

hat Jackson goes to the gym is 2.05. The probability that he goes 0 days is 0.1, the probability that he goes 2 days is , and the probability that he goes 3 days is .
Mathematics
2 answers:
Zina [9.1K]4 days ago
8 0
Jackson attends the gym either 0, 2, or 3 days weekly. Let’s denote P(0 days)=p, P(2 days)=q, and P(3 days)=r. Hence, the equation p+q+r equals 1. The expectation formula gives us E(X) = E(x=0)*P(0 days) + E(x=2)*P(2 days) + E(x=3)*P(3 days) or E(X) = 0*p + 2*q + 3*r = 2q + 3r. We're given P(0 days) = p = 0.1 and E(X) = 2.05, establishing the equations 0.1 + q + r = 1 and 2q + 3r = 2.05. These can be rearranged to q + r = 0.9 and 2q + 3r = 2.05. By solving these equations: (2q + 3r) - 2*(q + r) = 2.05 - 2(0.9) simplifies to r = 0.25, leading to q = 0.65. Thus, we find that P(2 days) = 0.65 and P(3 days) = 0.25.
Svet_ta [9.5K]4 days ago
6 0
In a probability distribution, the expected value is computed by summing the products of probabilities and their corresponding data values. Denote x as the probability of Jackson attending the gym for 2 days and y as the probability for 3 days. The probabilities and their values are as follows: 0 days: 0.1, 2 days: x, and 3 days: y. Therefore, the expected value can be expressed as: 0(0.1) + 2(x) + 3(y). We're given that the expected value is 2.05, leading to the equation 2x + 3y = 2.05 (Equation 1). Additionally, the total probability must equal 1, allowing us to set the second equation as: 0.1 + x + y = 1, which simplifies to x + y = 0.9 (Equation 2). By rearranging Equation 2, we substitute x as 0.9 - y into Equation 1, yielding: 2(0.9 - y) + 3y = 2.05, simplifying to 1.8 - 2y + 3y = 2.05, hence 1.8 + y = 2.05, inferring y = 0.25. Substituting this back into Equation 2 leads to x = 0.65. Finally, we conclude that the probability of Jackson going to the gym for 2 days is 0.65, while for 3 days is 0.25.
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Step-by-step explanation:

A frog positioned right at the center of a 5ft long board is 2.5 ft away from either edge.

Every 10 seconds, the frog jumps left or right.

If the frog's jumps are LLRLRL, it will remain on the board at the leftmost square.

If it jumps as LLRLL, it will jump off the board after fifty seconds.

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   (using digits from a random number table)

  • It requires 7 jumps with 2 in reverse direction (either left or right) for the frog to get off the board in 60 seconds.
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1 month ago
According to a survey by Bankrate, of adults in the United States save nothing for retirement (CNBC website). Suppose that adult
tester [8842]

Complete Question

The complete question appears in the first uploaded image

Answer:

a) Yes, selecting 15 corresponds to a binomial experiment

b)

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d) P(r = 15) = 3.2768 *10^{-11}

Step-by-step explanation:

Regarding question a:

For an experiment to qualify as binomial

the trials have to be independent

each trial must yield one of two possible outcomes

Given that the selection of 15 individuals is random, we ascertain that the trials are independent and the outcomes are “either the individual saves for retirement or does not save for retirement.”

Therefore, we conclude that the selection of 15 people at random is indeed a binomial experiment.

In question b:

The probability that all selected adults do not save for retirement is mathematically modeled as

P(r = n) = ^nC_r * p^r * q^{n-r}

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r = 15 implies all selected adults

n refers to the population size equating to 15

From the problem, p = 0.20

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=> q = 1 - p

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P(r = 15) = ^{15}C_{15} * p^{15} * q^{15-15}

P(r = 15) = 3.2768 *10^{-11} Regarding question c:

The probability that exactly five of the selected adults do not save for retirement is mathematically modeled as

P(r = 5) = ^{15} C_5 * (0.20)^5 * (0.80)^{15}

P(r = 5) = 0.1032

In relation to question d:

The probability that at least one of the selected adults opts not to save for retirement can be mathematically expressed as

P(r \ge 1 ) = 1 - P (r = 0 )

P(r \ge 1 ) = 1 - [ ^{15} C _ 0 * (0.20)^{0} * (0.80 )^{15}]

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14 days ago
Ann and Bill play rock-paper-scissors. Each has a strategy of choosing uniformly at random out of the three possibilities every
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Answer:

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c) The probability that Ann's first win occurs after the fourth round is 16/81

Step-by-step explanation:

a) Each strategy is played with a probability of 1/3. Given any strategy, there’s a 1/3 chance that Bill will choose the strategy that allows Ann to win. Consequently, the probability of Ann securing a victory in the first round (or any round) is

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Thus, the likelihood of Ann winning the initial round is 1/3.

b) The chances of Ann winning a round stand at 1/3; therefore, her chances of not winning are 2/3. This must happen three times before her first victory. Thus, the probability that Ann's first win occurs in the fourth round is

(2/3)³ * 1/3 = 8/81.

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