A random sample of size 100 is drawn from a population with a standard deviation of 10. If only 5% of the time a sample mean greater than 20 is obtained, the mean of the population is ______.
Z = (score-mean)/SEm
SEm = SD/√n
Assuming a normal distribution, find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.05) and its Z score. Insert Z score and other data into the top equation and solve for the mean.
Well, isn't this a fetching statistical problem! It's like playing hide and seek with numbers. So, we have a sample of size 100, and we know that only 5% of the time we get a sample mean greater than 20. If that's the case, then we can assume that the mean of the population must be, well, less than 20. Why? Because if it were greater, we'd have more than 5% of the time getting a mean greater than 20. So, let's keep it simple and say the mean of the population is less than...wait for it...20! There you have it, my friend, a dazzling answer!
To determine the mean of the population, we will use the concept of the standard error of the mean and the z-score.
The standard error of the mean (SE) is calculated as the standard deviation of the population divided by the square root of the sample size.
In this case, the standard deviation of the population is given as 10, and the sample size is 100. Thus, the standard error of the mean (SE) is calculated as:
SE = standard deviation / square root of sample size
SE = 10 / √100
SE = 10 / 10
SE = 1
To find the z-score, we need to convert the given percentage to a corresponding z-score. Since we want to find the sample mean greater than 20, which means the sample mean is in the top 5%, we need to find the z-score that corresponds to the top 5% of the standard normal distribution.
The z-score corresponding to the top 5% is 1.645.
Now, we can use the z-score formula to find the sample mean:
z-score = (sample mean - population mean) / standard error of the mean
Solving for the population mean:
1.645 = (20 - population mean) / 1
Rearranging the equation:
20 - population mean = 1.645
Lastly, solving for the population mean:
population mean = 20 - 1.645
population mean ≈ 18.355
Therefore, the mean of the population is approximately 18.355.
To find the mean of the population, we can use the concept of sampling distribution and the properties of the normal distribution.
First, let's understand the concept of a sampling distribution. A sampling distribution is the probability distribution of a statistic (in this case, the sample mean) for all possible samples of a fixed size drawn from a population. According to the Central Limit Theorem, when the sample size is large enough, the sampling distribution of the sample mean follows a normal distribution, regardless of the shape of the population distribution.
The given information states that the random sample of size 100 has a standard deviation of 10. Since the sample size is reasonably large, we can assume that the sampling distribution of the sample mean is normally distributed.
Now, let's consider the fact that only 5% of the time a sample mean greater than 20 is obtained. This means we are looking for the z-score corresponding to a cumulative probability of 0.95 (since we want to find the value that is below the top 5% of the distribution).
Using a standard normal distribution table or a statistical calculator, we can find that the z-score corresponding to a cumulative probability of 0.95 is approximately 1.645.
The formula to calculate the z-score is given by:
z = (x - μ) / (σ / sqrt(n))
Where:
z is the z-score
x is the value under consideration (20 in this case)
μ is the mean of the population (what we need to find)
σ is the standard deviation of the population (given as 10)
n is the sample size (given as 100)
Rearranging the formula to find μ, we get:
μ = x - (z * σ / sqrt(n))
= 20 - (1.645 * 10 / sqrt(100))
= 20 - (1.645 * 10 / 10)
= 20 - 1.645
= 18.355
Therefore, the mean of the population is approximately 18.355.