A final exam in Math 157 is normally distributed and has a mean of 75 with a standard deviation of 12. If 36 students are randomly selected, find the probability that the mean of their test scores is greater than 70.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)(but 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 proportion related to the Z score.
To find the probability that the mean of the test scores is greater than 70, we need to use the concept of the sampling distribution of the sample mean. In this case, we can treat the mean test scores of the 36 randomly selected students as a sample mean.
The sampling distribution of the sample mean follows a normal distribution when the sample size is sufficiently large and the population distribution is normal or nearly normal. In this case, we can assume that it is valid to use the normal distribution to approximate the sampling distribution of the sample mean.
The mean of the sampling distribution of the sample mean (μx̅) is equal to the population mean (μ), which in this case is 75. The standard deviation of the sampling distribution of the sample mean (σx̅) is equal to the population standard deviation (σ) divided by the square root of the sample size (n).
In this case, the standard deviation of the sampling distribution of the sample mean is:
σx̅ = σ / sqrt(n) = 12 / sqrt(36) = 12 / 6 = 2
Now, we want to find the probability that the mean of the test scores is greater than 70, which can be expressed as P(x̅ > 70). We can standardize this value using the standard deviation of the sampling distribution:
z = (x̅ - μx̅) / σx̅ = (70 - 75) / 2 = -2.5
Now, we need to find the area under the standard normal distribution curve to the right of z = -2.5, which represents the probability that the mean of the test scores is greater than 70. Using a standard normal distribution table or a calculator, we can find this probability.
The probability P(x̅ > 70) can be calculated using the standard normal distribution table or a calculator:
P(x̅ > 70) = P(z > -2.5)
Using a standard normal distribution table, we find that P(z > -2.5) is approximately 0.9938.
Therefore, the probability that the mean of the test scores is greater than 70 for a randomly selected sample of 36 students is approximately 0.9938, or 99.38%.