To solve this problem, we first need to understand the given information. We have a discrete uniform population with possible values of 2, 4, and 6. This means that each of these values has an equal probability of being chosen, which is 1/3.

Next, we need to calculate the mean (μ) and standard deviation (σ) of this population. The mean is the average of the possible values, which is (2+4+6)/3 = 4. The variance is the average of the squared differences from the Mean, which is [(2-4)²+(4-4)²+(6-4)²]/3 = 8/3. The standard deviation is the square root of the variance, which is √(8/3) ≈ 1.63.

We are selecting a sample of size 54 with replacement, which means that each selection is independent and has the same probability distribution. The Central Limit Theorem tells us that the distribution of the sample mean will be approximately normal with mean μ and standard deviation σ/√n, where n is the sample size. In this case, μ = 4 and σ/√n = 1.63/√54 ≈ 0.22.

We want to find the probability that the sample mean is greater than 4.1 but less than 4.4. To do this, we need to convert these values to z-scores, which measure how many standard deviations away from the mean they are. The z-score is calculated as (x - μ) / (σ/√n). For 4.1, the z-score is (4.1 - 4) / 0.22 ≈ 0.45. For 4.4, the z-score is (4.4 - 4) / 0.22 ≈ 1.82.

Finally, we use the standard normal distribution (z-distribution) to find the probabilities corresponding to these z-scores. The probability of a z-score being less than 0.45 is approximately 0.6736 and the probability of a z-score being less than 1.82 is approximately 0.9656. The probability that the z-score is between 0.45 and 1.82 is therefore 0.9656 - 0.6736 = 0.2920, or 29.20%.

So, the probability that a random sample of size 54, selected with replacement, will yield a sample mean greater than 4.1 but less than 4.4 is approximately 29.20%.

Question

To solve this problem, we first need to understand the given information. We have a discrete uniform population with possible values of 2, 4, and 6. This means that each of these values has an equal probability of being chosen, which is 1/3.

Next, we need to calculate the mean (μ) and standard deviation (σ) of this population. The mean is the average of the possible values, which is (2+4+6)/3 = 4. The variance is the average of the squared differences from the Mean, which is [(2-4)²+(4-4)²+(6-4)²]/3 = 8/3. The standard deviation is the square root of the variance, which is √(8/3) ≈ 1.63.

We are selecting a sample of size 54 with replacement, which means that each selection is independent and has the same probability distribution. The Central Limit Theorem tells us that the distribution of the sample mean will be approximately normal with mean μ and standard deviation σ/√n, where n is the sample size. In this case, μ = 4 and σ/√n = 1.63/√54 ≈ 0.22.

We want to find the probability that the sample mean is greater than 4.1 but less than 4.4. To do this, we need to convert these values to z-scores, which measure how many standard deviations away from the mean they are. The z-score is calculated as (x - μ) / (σ/√n). For 4.1, the z-score is (4.1 - 4) / 0.22 ≈ 0.45. For 4.4, the z-score is (4.4 - 4) / 0.22 ≈ 1.82.

Finally, we use the standard normal distribution (z-distribution) to find the probabilities corresponding to these z-scores. The probability of a z-score being less than 0.45 is approximately 0.6736 and the probability of a z-score being less than 1.82 is approximately 0.9656. The probability that the z-score is between 0.45 and 1.82 is therefore 0.9656 - 0.6736 = 0.2920, or 29.20%.

So, the probability that a random sample of size 54, selected with replacement, will yield a sample mean greater than 4.1 but less than 4.4 is approximately 29.20%.

Knowee AI · Accepted Answer

To solve this problem, we first need to understand the given information. We have a discrete uniform population with possible values of 2, 4, and 6. This means that each of these values has an equal probability of being chosen, which is 1/3.

Next, we need to calculate the mean (μ) and standard deviation (σ) of this population. The mean is the average of the possible values, which is (2+4+6)/3 = 4. The variance is the average of the squared differences from the Mean, which is [(2-4)²+(4-4)²+(6-4)²]/3 = 8/3. The standard deviation is the square root of the variance, which is √(8/3) ≈ 1.63.

We are selecting a sample of size 54 with replacement, which means that each selection is independent and has the same probability distribution. The Central Limit Theorem tells us that the distribution of the sample mean will be approximately normal with mean μ and standard deviation σ/√n, where n is the sample size. In this case, μ = 4 and σ/√n = 1.63/√54 ≈ 0.22.

We want to find the probability that the sample mean is greater than 4.1 but less than 4.4. To do this, we need to convert these values to z-scores, which measure how many standard deviations away from the mean they are. The z-score is calculated as (x - μ) / (σ/√n). For 4.1, the z-score is (4.1 - 4) / 0.22 ≈ 0.45. For 4.4, the z-score is (4.4 - 4) / 0.22 ≈ 1.82.

Finally, we use the standard normal distribution (z-distribution) to find the probabilities corresponding to these z-scores. The probability of a z-score being less than 0.45 is approximately 0.6736 and the probability of a z-score being less than 1.82 is approximately 0.9656. The probability that the z-score is between 0.45 and 1.82 is therefore 0.9656 - 0.6736 = 0.2920, or 29.20%.

So, the probability that a random sample of size 54, selected with replacement, will yield a sample mean greater than 4.1 but less than 4.4 is approximately 29.20%.

Given the discrete uniform population f(x) = 1 3 , x = 2, 4, 6, 0, elsewhere, find the probability that a random sample of size 54, selected with replacement, will yield a sample mean greater than 4.1 but less than 4.4. Assume the means are measured to the nearest tenth.

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