a population forms a normal distribution with a mean of u=80 and a standard deviation of o=15. for each of the following samples, compute the z-score for the sample mean and determine whether the sample mean is a typical, representative value or an extreme value for a sample of this size.
a. m=84 for n=9 scores
b. m=84 for n=100 scores
To compute the z-scores for the sample means, we can use the formula:
z = (x - μ) / (σ / √n)
where:
z is the z-score
x is the sample mean
μ is the population mean
σ is the population standard deviation
n is the sample size
a) For sample mean m = 84 and n = 9:
Using the formula, we have:
z = (84 - 80) / (15 / √9)
z = 4 / (15 / 3)
z = 4 / 5
z = 0.8
To determine whether the sample mean is a typical, representative value or an extreme value, we can refer to the standard z-score table. Since the z-score of 0.8 falls within the range of -1.96 and +1.96 (which encompasses approximately 95% of the data in a normal distribution), we can conclude that the sample mean of 84 is a typical, representative value for a sample of this size.
b) For sample mean m = 84 and n = 100:
Using the formula, we have:
z = (84 - 80) / (15 / √100)
z = 4 / (15 / 10)
z = 4 / 1.5
z ≈ 2.67
Similarly, referring to the standard z-score table, we find that the z-score of 2.67 falls outside the range of -1.96 and +1.96. Therefore, we can conclude that the sample mean of 84 is an extreme value for a sample of size 100.
In summary:
a) The sample mean of 84 for a sample size of 9 is a typical, representative value.
b) The sample mean of 84 for a sample size of 100 is an extreme value.
To compute the z-score for the sample mean, we can use the formula:
z = (x - μ) / (σ / √n)
where:
- x is the sample mean
- μ is the population mean
- σ is the population standard deviation
- n is the sample size
Let's calculate the z-scores for each case:
a. m = 84 for n = 9 scores:
Here, the sample mean is 84, the population mean is 80, the population standard deviation is 15, and the sample size is 9.
z = (84 - 80) / (15 / √9)
z = 4 / (15 / 3)
z = 4 / 5
z = 0.8
b. m = 84 for n = 100 scores:
In this case, the sample mean is still 84, the population mean and standard deviation remain 80 and 15, respectively, but the sample size is now 100.
z = (84 - 80) / (15 / √100)
z = 4 / (15 / 10)
z = 4 / 1.5
z = 2.67
Now, let's determine whether these z-scores represent typical or extreme values for samples of their respective sizes.
In a normal distribution:
- A z-score between -1 and 1 represents a typical value,
- A z-score less than -1 or greater than 1 represents an extreme value.
For case (a), with a z-score of 0.8, it falls within the range of -1 to 1, making it a typical value for a sample of size 9.
For case (b), with a z-score of 2.67, it is greater than 1, indicating it is an extreme value for a sample of size 100.
Therefore, the sample mean of 84 is considered representative for a sample of size 9, but it is considered an extreme value for a sample of size 100.