`The average grade point average (GPA) of undergraduate students in New York is normally distributed with a population mean of 2.5 and a population standard deviation of 0.4. Compute the following, showing all work:
1) The percentage of students with GPA's between 2.0 and 2.6
2) The percentage of students with GPA's below 2.7
3) Above what GPA will the top 7% of the students be(i.e., compute the 93rd percentile)
4) If a sample of 49 students is taken, what is the probability that the sample mean GPA will be between 2.60 and 2.70?
To solve these problems, we will use the standard normal distribution. Since we know the population mean and standard deviation, we can convert the given values to their corresponding z-scores, which will allow us to use the standard normal distribution table.
1) The percentage of students with GPAs between 2.0 and 2.6:
First, we find the z-scores corresponding to both GPA values:
z1 = (2.0 - 2.5) / 0.4 = -1.25
z2 = (2.6 - 2.5) / 0.4 = 0.25
Next, we need to find the area under the standard normal distribution curve between these two z-scores. By referring to the standard normal distribution table (or using a calculator/app that provides such functionality), we find the corresponding values:
P(z < -1.25) = 0.1056
P(z < 0.25) = 0.5987
To find the percentage of students with GPAs between 2.0 and 2.6, we calculate:
Percentage = (0.5987 - 0.1056) * 100
= 49.31%
Therefore, approximately 49.31% of students have GPAs between 2.0 and 2.6.
2) The percentage of students with GPAs below 2.7:
First, we need to find the z-score for GPA 2.7:
z = (2.7 - 2.5) / 0.4 = 0.5
Now, we find the area under the standard normal distribution curve to the left of this z-score:
P(z < 0.5) = 0.6915
Therefore, approximately 69.15% of students have GPAs below 2.7.
3) The GPA above which the top 7% of students will be:
We are looking for the GPA value that corresponds to the 93rd percentile. We need to find the z-score that corresponds to this percentile:
P(Z > z) = 0.07
From the standard normal distribution table, we find the z-score for a cumulative probability of 0.93, which is approximately 1.48.
Now, we can solve for the corresponding GPA value:
z = (GPA - 2.5) / 0.4
1.48 = (GPA - 2.5) / 0.4
Simplifying this equation:
0.592 = GPA - 2.5
GPA = 2.5 + 0.592
GPA = 3.092
Therefore, the GPA above which the top 7% of students will be is approximately 3.092.
4) The probability that the sample mean GPA will be between 2.60 and 2.70 (assuming a sample size of 49):
Since we are dealing with the sample mean GPA, we will use the concept of the sampling distribution of the mean. According to the Central Limit Theorem, as long as the sample size is large enough (typically above 30), the sampling distribution of the mean will be approximately normally distributed, regardless of the shape of the population distribution.
To find the probability, we need to calculate the z-scores for both GPA values:
z1 = (2.60 - 2.5) / (0.4 / sqrt(49)) = 1.23
z2 = (2.70 - 2.5) / (0.4 / sqrt(49)) = 1.73
Now, we need to find the area under the standard normal distribution curve between these two z-scores. Using the standard normal distribution table or a calculator/app, we find:
P(1.23 < z < 1.73) = P(z < 1.73) - P(z < 1.23)
= 0.9571 - 0.8915
= 0.0656
Therefore, the probability that the sample mean GPA is between 2.60 and 2.70 is approximately 0.0656, or 6.56%.