To solve this problem, we can use the Central Limit Theorem which states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large (usually n 30). However, for non-normal populations, the sample size can be less than 30, if the population is not heavily skewed or has no extreme outliers.

In this case, we have a sample size of 17, which is less than 30, but the problem states that the distribution is left-skewed, not heavily skewed or with extreme outliers. So, we can still use the Central Limit Theorem.

The mean (μ) of the population is 63.78 years and the standard deviation (σ) is 3.89 years.

We want to find the probability that the sample mean is at most 63.88 years.

First, we need to find the standard deviation of the sample mean (σx̄), which is equal to the standard deviation of the population (σ) divided by the square root of the sample size (n).

σx̄ = σ/√n = 3.89/√17 ≈ 0.943

Next, we find the z-score, which is the number of standard deviations a particular score is from the mean. The z-score is calculated as:

z = (x̄ - μ) / σx̄

where x̄ is the sample mean, μ is the population mean, and σx̄ is the standard deviation of the sample mean.

z = (63.88 - 63.78) / 0.943 ≈ 0.106

Finally, we use a z-table to find the probability that the z-score is less than or equal to 0.106. Looking up this z-score in the z-table, we find a probability of approximately 0.5423.

So, the probability that the sample has an average value of at most 63.88 years is approximately 0.5423, or 54.23%.

Therefore, the answer is 0.5423.

Question

To solve this problem, we can use the Central Limit Theorem which states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large (usually n > 30). However, for non-normal populations, the sample size can be less than 30, if the population is not heavily skewed or has no extreme outliers.

In this case, we have a sample size of 17, which is less than 30, but the problem states that the distribution is left-skewed, not heavily skewed or with extreme outliers. So, we can still use the Central Limit Theorem.

The mean (μ) of the population is 63.78 years and the standard deviation (σ) is 3.89 years.

We want to find the probability that the sample mean is at most 63.88 years.

First, we need to find the standard deviation of the sample mean (σx̄), which is equal to the standard deviation of the population (σ) divided by the square root of the sample size (n).

σx̄ = σ/√n = 3.89/√17 ≈ 0.943

Next, we find the z-score, which is the number of standard deviations a particular score is from the mean. The z-score is calculated as:

z = (x̄ - μ) / σx̄

where x̄ is the sample mean, μ is the population mean, and σx̄ is the standard deviation of the sample mean.

z = (63.88 - 63.78) / 0.943 ≈ 0.106

Finally, we use a z-table to find the probability that the z-score is less than or equal to 0.106. Looking up this z-score in the z-table, we find a probability of approximately 0.5423.

So, the probability that the sample has an average value of at most 63.88 years is approximately 0.5423, or 54.23%.

Therefore, the answer is 0.5423.

Knowee AI · Accepted Answer

To solve this problem, we can use the Central Limit Theorem which states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large (usually n > 30). However, for non-normal populations, the sample size can be less than 30, if the population is not heavily skewed or has no extreme outliers.