To solve this problem, we need to use the concept of the sampling distribution of the sample mean.

Step 1: Identify the parameters of the population distribution
The problem tells us that the population has a mean (μ) of 33 and a standard deviation (σ) of 6.

Step 2: Identify the parameters of the sampling distribution
The mean of the sampling distribution (μx̄) is equal to the population mean, which is 33. The standard deviation of the sampling distribution (σx̄), also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size (n). In this case, n = 100, so σx̄ = σ/√n = 6/√100 = 0.6.

Step 3: Standardize the desired sample mean range
We want to find the probability that the sample mean is between 34.5 and 36.1. To do this, we first convert these values to z-scores, which are measures of how many standard errors the values are away from the mean of the sampling distribution. The z-scores are calculated as follows:

z1 = (x̄1 - μx̄)/σx̄ = (34.5 - 33)/0.6 = 2.5
z2 = (x̄2 - μx̄)/σx̄ = (36.1 - 33)/0.6 = 5.17

Step 4: Find the probability associated with the z-scores
We want to find the probability that the z-score is between 2.5 and 5.17. Looking up these z-scores on a standard normal (Z) table or using a Z-table calculator, we find that the area to the left of z1 = 2.5 is approximately 0.9938 and the area to the left of z2 = 5.17 is essentially 1 (because 5.17 is so far out in the tail of the distribution that the area to its left is virtually the entire distribution).

The probability that the z-score is between 2.5 and 5.17 is the area between these z-scores, which is the area to the left of z2 minus the area to the left of z1. So, the probability is 1 - 0.9938 = 0.0062.

So, the probability that the sample mean of 100 observations taken at random from this population will be between 34.5 and 36.1 is approximately 0.0062. Therefore, the correct answer is 0.0062.

Question

To solve this problem, we need to use the concept of the sampling distribution of the sample mean.

Step 1: Identify the parameters of the population distribution
The problem tells us that the population has a mean (μ) of 33 and a standard deviation (σ) of 6.

Step 2: Identify the parameters of the sampling distribution
The mean of the sampling distribution (μx̄) is equal to the population mean, which is 33. The standard deviation of the sampling distribution (σx̄), also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size (n). In this case, n = 100, so σx̄ = σ/√n = 6/√100 = 0.6.

Step 3: Standardize the desired sample mean range
We want to find the probability that the sample mean is between 34.5 and 36.1. To do this, we first convert these values to z-scores, which are measures of how many standard errors the values are away from the mean of the sampling distribution. The z-scores are calculated as follows:

z1 = (x̄1 - μx̄)/σx̄ = (34.5 - 33)/0.6 = 2.5
z2 = (x̄2 - μx̄)/σx̄ = (36.1 - 33)/0.6 = 5.17

Step 4: Find the probability associated with the z-scores
We want to find the probability that the z-score is between 2.5 and 5.17. Looking up these z-scores on a standard normal (Z) table or using a Z-table calculator, we find that the area to the left of z1 = 2.5 is approximately 0.9938 and the area to the left of z2 = 5.17 is essentially 1 (because 5.17 is so far out in the tail of the distribution that the area to its left is virtually the entire distribution).

The probability that the z-score is between 2.5 and 5.17 is the area between these z-scores, which is the area to the left of z2 minus the area to the left of z1. So, the probability is 1 - 0.9938 = 0.0062.

So, the probability that the sample mean of 100 observations taken at random from this population will be between 34.5 and 36.1 is approximately 0.0062. Therefore, the correct answer is 0.0062.

Knowee AI · Accepted Answer