Essentials of statistics is the subject and the question is: The population of IQ scores forms a normal distribution with a mean of u=100 and a standard deviation of sigma=15. What is the probability of obtaining a sample mean greater than M=105 for a random sample of n=36 people? What is the formula in simple terms.
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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 corresponding the the Z score.
To find the probability of obtaining a sample mean greater than M=105 for a random sample of n=36 people, we need to utilize the central limit theorem and z-scores.
The central limit theorem states that the distribution of sample means for a large enough sample size will follow a normal distribution, regardless of the shape of the population distribution. In this case, since the population of IQ scores is normally distributed, the sample means will also be normally distributed.
To calculate the z-score for a sample mean, we 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
In this case, X = 105, μ = 100, σ = 15, and n = 36. Plugging in these values, we can calculate the z-score:
z = (105 - 100) / (15 / √36)
z = 5 / (15 / 6)
z = 5 / 2.5
z = 2
Next, we need to calculate the probability of obtaining a z-score greater than 2. We can refer to the z-table (a table of standard normal distribution) or use a statistical software to find this probability.
Using the z-table, we can look up the area to the right of z = 2. The table will give us the probability associated with that value. Let's assume the probability is P.
Therefore, the probability of obtaining a sample mean greater than M=105 for a random sample of n=36 people is P.