A random sample of n= 16 scores is selected from a normal distribution with a mean of u=50 and a standard deviation of q=10. A. what is th probability that the sample mean will have a value between 45 and 55? B. What is the probability that a sample mean will have a value between 488 and 52? C. What range of value has a 95% probability of containing the sample mean?
Z = (mean1 - mean2)/standard error (SE) of difference between means
mean1 = sample mean
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1) (You can use n)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores you found.
To answer these questions, we need to use the concept of the sampling distribution of the sample mean and the Central Limit Theorem. The Central Limit Theorem states that, for a large enough sample size (n), the sampling distribution of the sample mean will follow a normal distribution, regardless of the shape of the population distribution.
The mean of the sampling distribution of the sample mean is equal to the population mean (u), and the standard deviation is equal to the population standard deviation (q) divided by the square root of the sample size (sqrt(n)).
In this case, the mean (u) is 50 and the standard deviation (q) is 10. The sample size (n) is 16. Substituting these values into the formula, we can find the standard deviation of the sampling distribution (denoted as "σ x̄ "):
σ x̄ = q / sqrt(n)
σ x̄ = 10 / sqrt(16) = 10 / 4 = 2.5
Now let's answer each part of the question:
A. What is the probability that the sample mean will have a value between 45 and 55?
To calculate this probability, we need to convert these values to z-scores (standardized scores). The formula to calculate the z-score is:
z = (x - u) / σ x̄
For x = 45:
z1 = (45 - 50) / 2.5 = -2
For x = 55:
z2 = (55 - 50) / 2.5 = 2
Next, we need to find the area under the standard normal curve between these two z-scores. We can use a Z-table or a calculator to find the probabilities associated with these z-scores.
Looking up the probabilities corresponding to z = -2 and z = 2 in a standard normal distribution table, we find:
P(z < -2) ≈ 0.0228
P(z < 2) ≈ 0.9772
To find the probability between -2 and 2 (45 and 55), we subtract the smaller probability from the larger probability:
P(-2 < z < 2) ≈ 0.9772 - 0.0228 ≈ 0.9544
So, the probability that the sample mean will have a value between 45 and 55 is approximately 0.9544, or 95.44%.
B. What is the probability that a sample mean will have a value between 48 and 52?
Using the same process as above, we can calculate the z-scores:
For x = 48:
z1 = (48 - 50) / 2.5 = -0.8
For x = 52:
z2 = (52 - 50) / 2.5 = 0.8
Again, looking up the probabilities corresponding to z = -0.8 and z = 0.8 in the standard normal distribution table, we find:
P(z < -0.8) ≈ 0.2119
P(z < 0.8) ≈ 0.7881
To find the probability between -0.8 and 0.8 (48 and 52), we subtract the smaller probability from the larger probability:
P(-0.8 < z < 0.8) ≈ 0.7881 - 0.2119 ≈ 0.5762
So, the probability that the sample mean will have a value between 48 and 52 is approximately 0.5762, or 57.62%.
C. What range of values has a 95% probability of containing the sample mean?
To find the range of values that has a 95% probability of containing the sample mean, we need to find the z-scores associated with the tails of the distribution that correspond to a cumulative probability of 2.5% on each side.
In a standard normal distribution, we need to find the z-scores that enclose 2.5% in both tails. We can find these z-scores from the Z-table or use the inverse normal distribution function of a calculator.
The z-score corresponding to a cumulative probability of 2.5% is approximately -1.96 in the left tail and 1.96 in the right tail.
Now, we can calculate the corresponding sample values using the formula:
x1 = u + z1 * σ x̄
x2 = u + z2 * σ x̄
For the left tail (lower bound):
x1 = 50 + (-1.96) * 2.5 ≈ 45.1
For the right tail (upper bound):
x2 = 50 + 1.96 * 2.5 ≈ 54.9
Therefore, the range of values that has a 95% probability of containing the sample mean is approximately from 45.1 to 54.9.