Question
The GPA at a particular school has an average of 2.89 with a standard deviation of 0.63. A random sample of 38 students from that school is collected. Find the probability that the average GPA for this sample is greater than 3.0.
Answers
GPT 3.5
To solve this problem, we will use the Central Limit Theorem, which states that the sampling distribution of the sample mean approaches a normal distribution with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
First, we calculate the standard error of the sample mean:
standard error = standard deviation / sqrt(sample size)
standard error = 0.63 / sqrt(38) = 0.102
Next, we calculate the z-score for the sample mean:
z = (sample mean - population mean) / standard error
z = (3.0 - 2.89) / 0.102 = 1.078
Using a standard normal distribution table or calculator, we can find the probability associated with a z-score of 1.078. This probability represents the area under the curve to the right of the z-score.
The probability that the average GPA for this sample is greater than 3.0 is approximately 0.1401, or 14.01%.
First, we calculate the standard error of the sample mean:
standard error = standard deviation / sqrt(sample size)
standard error = 0.63 / sqrt(38) = 0.102
Next, we calculate the z-score for the sample mean:
z = (sample mean - population mean) / standard error
z = (3.0 - 2.89) / 0.102 = 1.078
Using a standard normal distribution table or calculator, we can find the probability associated with a z-score of 1.078. This probability represents the area under the curve to the right of the z-score.
The probability that the average GPA for this sample is greater than 3.0 is approximately 0.1401, or 14.01%.