Given a normal population whise mean is 50 and whise standard deviation is 5, find the probability that a random sample of 25 has a mean greater than 52.
Z = (52-50)/SEm
SEm = SD/sqrt n
Look up Z in table in back of your text labeled area under normal distribution.
0.9772
To find the probability that a random sample of 25 has a mean greater than 52, we need to use the sampling distribution of the sample means, also known as the central limit theorem.
The central limit theorem states that for a large sample size, the sampling distribution of the sample means will be approximately normally distributed, regardless of the shape of the original population. The mean of the sampling distribution will be equal to the population mean, and the standard deviation of the sampling distribution (also known as the standard error) will be equal to the population standard deviation divided by the square root of the sample size.
In this case, the population mean is 50 and the population standard deviation is 5. Since our sample size is 25, the standard error will be equal to 5 divided by the square root of 25, which is 1.
We can now calculate the z-score for the sample mean of 52. The z-score measures the number of standard deviations a data point is from the mean. It is calculated by subtracting the population mean from the sample mean and then dividing by the standard error.
z = (sample mean - population mean) / standard error
= (52 - 50) / 1
= 2
Next, we need to find the area under the normal curve to the right of the z-score of 2. This will give us the probability that a random sample of 25 has a mean greater than 52.
Using a standard normal distribution table or a statistical software, we can find that the area to the right of a z-score of 2 is approximately 0.0228.
So the probability that a random sample of 25 has a mean greater than 52 is approximately 0.0228, or 2.28%.