A random sample is drawn from a population with mean μ = 68 and standard deviation σ = 5.7. [You may find it useful to reference the z table.]
a. Is the sampling distribution of the sample mean with n = 16 and n = 41 normally distributed? (Round the standard error to 3 decimal places.)
b. Can you conclude that the sampling distribution of the sample mean is normally distributed for both sample sizes?
multiple choice 1
Yes, both the sample means will have a normal distribution.
No, both the sample means will not have a normal distribution.
No, only the sample mean with n = 16 will have a normal distribution.
No, only the sample mean with n = 41 will have a normal distribution.
c. If the sampling distribution of the sample mean is normally distributed with n = 16, then calculate the probability that the sample mean falls between 68 and 71. (If appropriate, round final answer to 4 decimal places.)
multiple choice 2
We cannot assume that the sampling distribution of the sample mean is normally distributed.
We can assume that the sampling distribution of the sample mean is normally distributed and the probability that the sample mean falls between 68 and 71 is
d. If the sampling distribution of the sample mean is normally distributed with n = 41, then calculate the probability that the sample mean falls between 68 and 71. (If appropriate, round final answer to 4 decimal places.)
multiple choice 3
We cannot assume that the sampling distribution of the sample mean is normally distributed.
We can assume that the sampling distribution of the sample mean is normally distributed and the probability that the sample mean falls between 68 and 71 is
a. To determine if the sampling distribution of the sample mean is normally distributed, we need to check if the sample size is large enough for the Central Limit Theorem to apply. The Central Limit Theorem states that as the sample size increases, the sampling distribution of the sample mean becomes more and more normally distributed, regardless of the shape of the population distribution.
In this case, n = 16 and n = 41. According to the Central Limit Theorem, a sample size of 30 or larger is typically considered large enough for the sampling distribution of the sample mean to be considered approximately normal.
Therefore, with n = 16 and n = 41, we can conclude that the sampling distribution of the sample mean is normally distributed for both sample sizes.
b. The answer is: Yes, both the sample means will have a normal distribution.
c. To calculate the probability that the sample mean falls between 68 and 71 with a normally distributed sampling distribution, we need to first calculate the z-scores for each value using the formula:
z = (x - μ) / (σ / sqrt(n))
For x = 68, μ = 68, σ = 5.7, and n = 16:
z1 = (68 - 68) / (5.7 / sqrt(16)) = 0
For x = 71:
z2 = (71 - 68) / (5.7 / sqrt(16)) = 1.403
Next, we need to find the corresponding probabilities from the z-table. The probability that the sample mean falls between 68 and 71 is given by the difference in these probabilities:
P(68 < x < 71) = P(0 < z < 1.403)
Using the z-table, we find that P(0 < z < 1.403) is approximately 0.9192.
Therefore, the probability that the sample mean falls between 68 and 71 is approximately 0.9192.
d. Following similar steps as in part c, we calculate the z-scores for x = 68 and x = 71 with n = 41:
For x = 68:
z1 = (68 - 68) / (5.7 / sqrt(41)) = 0
For x = 71:
z2 = (71 - 68) / (5.7 / sqrt(41)) = 1.929
Using the z-table, we find that P(0 < z < 1.929) is approximately 0.9738.
Therefore, the probability that the sample mean falls between 68 and 71 is approximately 0.9738.
a. To determine if the sampling distribution of the sample mean with n=16 and n=41 is normally distributed, we need to check if the sample size is large enough. According to the Central Limit Theorem, if the sample size is at least 30 or the population distribution is approximately normal, then the sampling distribution of the sample mean will be approximately normal.
For n=16, the sample size is less than 30, so we need to check if the population distribution is approximately normal. Since the question does not provide information about the population distribution, we cannot determine if the sampling distribution of the sample mean with n=16 is normally distributed.
For n=41, the sample size is greater than or equal to 30, so we can approximate the sampling distribution of the sample mean to be normally distributed.
b. The correct answer is: No, only the sample mean with n=41 will have a normal distribution.
c. To calculate the probability that the sample mean falls between 68 and 71, we need to use the z-score formula:
z = (x - μ) / (σ / √n)
Where:
x = sample mean (68)
μ = population mean (68)
σ = population standard deviation (5.7)
n = sample size (16)
First, let's calculate the z-scores:
z1 = (68 - 68) / (5.7 / √16) = 0 / 1.425 = 0
z2 = (71 - 68) / (5.7 / √16) = 3 / 1.425 = 2.105
Next, we need to find the cumulative probability associated with these z-scores using a z-table.
From the z-table, we find that the cumulative probability for z = 0 is 0.5000 and the cumulative probability for z = 2.105 is 0.9833.
To find the probability that the sample mean falls between 68 and 71, we subtract the cumulative probabilities:
Probability = 0.9833 - 0.5000 = 0.4833
The probability that the sample mean falls between 68 and 71 is approximately 0.4833.
d. Similarly, for n = 41, we can use the z-score formula:
z = (x - μ) / (σ / √n)
Where:
x = sample mean (68)
μ = population mean (68)
σ = population standard deviation (5.7)
n = sample size (41)
Calculating the z-scores:
z1 = (68 - 68) / (5.7 / √41) = 0 / 0.888 = 0
z2 = (71 - 68) / (5.7 / √41) = 3 / 0.888 = 3.377
Using the z-table, the cumulative probability for z = 0 is 0.5000 and the cumulative probability for z = 3.377 is 0.9995.
To find the probability that the sample mean falls between 68 and 71, we subtract the cumulative probabilities:
Probability = 0.9995 - 0.5000 = 0.4995
The probability that the sample mean falls between 68 and 71 is approximately 0.4995.
To answer these questions, we need to understand the concept of the sampling distribution of the sample mean and how it relates to the population mean and standard deviation.
The sampling distribution of the sample mean refers to the distribution of all possible sample means that could be obtained from repeated sampling from a population. It is important to note that the shape of the sampling distribution is not influenced by the sample size (n), but rather by the shape of the population distribution itself.
a. To determine if the sampling distribution of the sample mean with n = 16 and n = 41 is normally distributed, we need to check if the sample size is large enough for the Central Limit Theorem (CLT) to apply. The CLT states that for a sample size of n, the sampling distribution of the sample mean will be approximately normally distributed if the population distribution is approximately normal or if the sample size is sufficiently large.
For n = 16, if the population distribution is not too skewed, the sampling distribution of the sample mean can be reasonably approximated to be normally distributed. The standard error (SE) of the sample mean is calculated by dividing the population standard deviation (σ) by the square root of the sample size (n). In this case, SE = σ / √n = 5.7 / √16 = 5.7 / 4 = 1.425. So, the standard error for n = 16 is approximately 1.425.
For n = 41, the sample size is larger, making it more likely that the sampling distribution of the sample mean will be normally distributed, regardless of the shape of the population distribution. The standard error for n = 41 is calculated in the same way as for n = 16, resulting in approximately 0.887.
b. Based on the information above, we can conclude that both the sample means will have a normal distribution. Therefore, the correct option is: Yes, both the sample means will have a normal distribution.
c. To calculate the probability that the sample mean falls between 68 and 71, we need to convert these values to standard units using the z-score formula: z = (x - μ) / SE. In this case, the sample mean is equal to 68. The z-score for 68 is: (68 - 68) / 1.425 = 0.
Next, we need to find the z-score for 71: (71 - 68) / 1.425 = 2.105.
Using the z-table or a statistical software, we can find the probability associated with a z-score of 2.105. The probability that the sample mean falls between 68 and 71 is equal to the area under the standard normal curve between these two z-scores. Look up the area for z = 2.105 in the z-table to find this probability.
d. The same steps can be applied to calculate the probability that the sample mean falls between 68 and 71 for a sample size of n = 41. Using the z-score formula with the standard error for n = 41 (0.887), we can compute the z-scores for 68 and 71. Then, we find the corresponding probabilities using the z-table or statistical software.