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.
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%.
To find the probability that the average GPA for the sample is greater than 3.0, we can use the Central Limit Theorem, which states that the distribution of sample means approximates a normal distribution, regardless of the shape of the population.
Step 1: Convert the problem to a standard normal distribution.
We need to calculate the z-score for the sample mean using the formula:
z = (x - μ) / (σ / √n)
where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x = 3.0, μ = 2.89, σ = 0.63, and n = 38.
z = (3.0 - 2.89) / (0.63 / √38)
= 0.11 / (0.63 / 6.16)
= 0.11 / 0.102
Step 2: Calculate the probability using the z-score.
We need to find the probability that the z-score is greater than the calculated value.
P(z > 0.11 / 0.102) = P(z > 1.08)
Using a standard normal distribution table or calculator, we find that P(z > 1.08) is approximately 0.1401.
Step 3: Interpret the result.
The probability that the average GPA for this sample is greater than 3.0 is approximately 0.1401, or 14.01%.
To find the probability that the average GPA for this sample is greater than 3.0, we need to use the Central Limit Theorem and the z-score formula.
Step 1: Convert the population mean and standard deviation to the distribution of sample means.
The mean of the sample means (μm) is equal to the population mean (μ) which is 2.89.
The standard deviation of the sample means (σm) is equal to the population standard deviation (σ) divided by the square root of the sample size (n), which is 0.63/√38.
Step 2: Calculate the z-score.
The z-score formula is given by: z = (x - μm) / σm
Where x is the desired value (3.0) and μm is the mean of the sample means, and σm is the standard deviation of the sample means.
Plugging in the values, we get: z = (3.0 - 2.89) / (0.63/√38)
Step 3: Find the probability using the z-table.
With the calculated z-score, we can look up the corresponding probability in the standard normal table (z-table) or use a statistical calculator. We are interested in finding the probability of obtaining a z-score greater than the calculated value.
Using the z-table, we can find the cumulative probability associated with the z-score. Subtract this cumulative probability from 1 to get the probability that the average GPA for this sample is greater than 3.0.
Alternatively, you can use a statistical calculator or software to directly find the probability associated with the z-score without having to use the z-table.
Note: The Central Limit Theorem assumes that the sample size is large enough (in general, n >= 30) for the sampling distribution of the sample mean to be approximately normal. In this case, with a sample size of 38, we can apply the Central Limit Theorem.