Assume we draw a simple random sample from a population having a mean of 100 and a standard deviation of 16. What is the probability that a sample mean will be within plus-or-minus two of the population mean for each of the following sample sizes?
n = 50
n=100
n=200
n=400
The probability of being within Z = ±2 for a normal distribution always approximates 95%. However, what you might be seeking is the high and low values that cut off that proportion.
Standard error of the mean (SE) = standard deviation/square root of n
Z = (raw score - mean)/SE
±2 = (raw score - 100)/SE
Use the four SEs to find the raw score (±).
I hope this helps.
I don't understand how to get the answer using the formula.
For n = 100:
Lower limit: 100 - 2*16/sqrt(100) = 96.8
Upper limit: 100 + 2*16/sqrt(100) = 103.2
I can't get the answer for the other ones
Can you please help me solve for N
To answer this question, we need to use the concept of the standard error of the mean (SEM) and the Central Limit Theorem (CLT).
The standard error of the mean (SEM) is calculated by dividing the standard deviation of the population by the square root of the sample size.
The formula for the SEM is:
SEM = σ / sqrt(n)
Where:
- σ is the population standard deviation
- n is the sample size
The Central Limit Theorem states that for a large enough sample size, the distribution of the sample means will be approximately normally distributed, regardless of the shape of the population distribution.
Now let's calculate the probabilities:
For n = 50:
SEM = 16 / sqrt(50) = 2.26
To find the probability that the sample mean will be within plus-or-minus two of the population mean, we need to find the area under the normal curve between (100 - 2) and (100 + 2) when the standard deviation is 2.26.
Using a standard normal distribution table or a statistical calculator, we can find that the approximate probability is 0.9545, or 95.45%.
For n = 100:
SEM = 16 / sqrt(100) = 1.6
Following the same process as above, the probability is approximately 0.9545, or 95.45%.
For n = 200:
SEM = 16 / sqrt(200) = 1.13
The probability is approximately 0.9545, or 95.45%.
For n = 400:
SEM = 16 / sqrt(400) = 0.8
The probability is approximately 0.9545, or 95.45%.
In summary, regardless of the sample size, the probability that the sample mean will be within plus-or-minus two of the population mean is approximately 0.9545, or 95.45%. This is because of the Central Limit Theorem, which ensures that the sample means will be normally distributed even if the population distribution is not.