The GPAs of all students enrolled at a large university have an approximately normal distribution with a mean of 3.02 and a standard deviation of .29. Find the probability that the mean GPA of a random sample of 20 students selected from this university is
2.87 or lower
Find z-score:
z = (2.87 - 3.02)/(.29/√20) = ?
Calculate.
Check a z-table for the probability using the z-score. (Remember that the question is asking for "2.87 or lower" when looking at the table.)
The GPAs of all students enrolled at a university have an approximately normal distribution with a mean of 3.02 and a standard deviation of 0.29.Calculate the probability that a student chosen at random would obtain the GPA less than 3.10.
Why did the GPA cross the road? Because it wanted to get to the other side... of the distribution!
To find the probability that the mean GPA of a random sample of 20 students is 2.87 or lower, we can use the central limit theorem. This theorem tells us that if we take a large enough sample size, the sampling distribution of the mean will be approximately normal, regardless of the shape of the population distribution.
In this case, the mean GPA of the population is 3.02 with a standard deviation of 0.29. Since we don't know the shape of the population distribution, we can assume that it's approximately normal.
To find the probability, we need to standardize the sample mean using the z-score formula:
z = (x - μ) / (σ / sqrt(n))
Where:
x = 2.87 (the mean we're interested in)
μ = 3.02 (mean of the population)
σ = 0.29 (standard deviation of the population)
n = 20 (sample size)
Substituting the values into the formula, we get:
z = (2.87 - 3.02) / (0.29 / sqrt(20))
Calculating this, we get:
z ≈ -2.19
Now we need to find the probability of z being less than or equal to -2.19. We can look this up in a standard normal distribution table or use a calculator. Using a calculator, we find that the probability is approximately 0.0142, or 1.42%.
So, the probability that the mean GPA of a random sample of 20 students is 2.87 or lower is about 0.0142 or 1.42%.
To find the probability that the mean GPA of a random sample of 20 students is 2.87 or lower, we need to use the Central Limit Theorem.
The Central Limit Theorem states that if the sample size is large enough, regardless of the shape of the population distribution, the distribution of the sample means will be approximately normally distributed.
In this case, we are given that the population GPA has an approximately normal distribution with a mean of 3.02 and a standard deviation of 0.29. Therefore, we can assume that the distribution of the sample means will also be approximately normal, with the same mean of 3.02 and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
To calculate the standard deviation of the sample means, we divide the population standard deviation by the square root of the sample size:
standard deviation of sample means = population standard deviation / √(sample size)
= 0.29 / √(20)
= 0.29 / 4.47
≈ 0.065
Now, we can use the Z-score formula to find the probability that the mean GPA of the sample is 2.87 or lower. The Z-score formula is given by:
Z = (X - μ) / σ
Where:
Z is the Z-score
X is the value we are interested in
μ is the mean of the population
σ is the standard deviation of the population (in this case, the standard deviation of the sample means)
So, in this case, we want to find the Z-score when the mean GPA is 2.87:
Z = (2.87 - 3.02) / 0.065
≈ -2.31
Now, we use a Z-table or calculator to find the cumulative probability associated with the Z-score of -2.31. The cumulative probability represents the probability that a Z-score is equal to or less than a given value.
Using a Z-table or calculator, we find that the cumulative probability associated with a Z-score of -2.31 is approximately 0.0107.
Therefore, the probability that the mean GPA of a random sample of 20 students selected from this university is 2.87 or lower is approximately 0.0107, or 1.07%.