for a sample of n = 36, find the probability of a sample mean being less than 12.2 if u = 12 and o = 0.95
To find the probability of a sample mean being less than 12.2, we need to use the z-score formula and the standard normal distribution.
Step 1: Calculate the standard error (SE) using the formula:
SE = σ/√n
where σ is the population standard deviation, and n is the sample size.
Given that σ = 0.95 (o in the question) and n = 36:
SE = 0.95/√36
SE = 0.95/6
SE = 0.1583 (rounded to four decimal places)
Step 2: Calculate the z-score using the formula:
z = (x - μ) / SE
where x is the sample mean, μ is the population mean, and SE is the standard error.
Given that x = 12.2 (sample mean) and μ = 12 (population mean):
z = (12.2 - 12) / 0.1583
z = 0.2 / 0.1583
z = 1.2634 (rounded to four decimal places)
Step 3: Identify the probability associated with the z-score using a standard normal distribution table or a calculator.
Looking up the z-score of 1.2634 in the standard normal distribution table, we find that the area to the left of the z-score is 0.8962 (rounded to four decimal places).
Therefore, the probability of a sample mean being less than 12.2 is 0.8962 (or 89.62% when expressed as a percentage).
To find the probability of a sample mean being less than 12.2, you can use the formula for the Z-score and the standard normal distribution.
The formula for the Z-score is given by:
Z = (x - μ) / (σ / √n)
Where:
- Z is the Z-score
- x is the sample mean (in this case, 12.2)
- μ is the population mean (u = 12)
- σ is the population standard deviation (o = 0.95)
- n is the sample size (n = 36)
First, calculate the standard error (SE) using the formula:
SE = σ / √n
SE = 0.95 / √36
SE ≈ 0.1583
Next, calculate the Z-score:
Z = (12.2 - 12) / 0.1583
Z ≈ 1.2616
Now, use a standard normal distribution table or a Z-score calculator to find the probability (p) associated with this Z-score of 1.2616.
The probability can be found by looking up the Z-score in the Z-table or by using a Z-score calculator. For example, using the Z-table, you would find that the probability associated with a Z-score of 1.2616 is approximately 0.8968.
So, the probability of a sample mean being less than 12.2 is approximately 0.8968, or 89.68%.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since 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 proportion related to that Z score.