solve The following data represents the GPA of a sample of 15 students enrolled in a math class. Round answers to at least 4 decimal places. For Excel quartiles use the =QUARTILE.EXC function.
GPA
2.26
3.01
3.02
2.54
3.32
3.14
3.82
2.39
3.04
2.3
2.7
3.52
2.75
2.38
3.87
(a) Find the first quartile?
(b) Find the third quartile?
(c) Find the mean?
(d) Find the median?
(e) Find the range?
(f) Find the standard deviation?
(g) Find the interquartile range (IQR)?
(h) What are the lower and upper limits to find outliers?
LL: UL:

Let's solve each part of the problem using the given data set.

### Given Data:
GPA: 
2.26, 3.01, 3.02, 2.54, 3.32, 3.14, 3.82, 2.39, 3.04, 2.3, 2.7, 3.52, 2.75, 2.38, 3.87

### Step-by-Step Solution:

#### (a) First Quartile (Q1)
To find the first quartile, we will arrange the data in ascending order and then apply the formula for the first quartile using Excel's `=QUARTILE.EXC` function.

Ordered Data:  
2.26, 2.30, 2.38, 2.39, 2.54, 2.70, 2.75, 3.01, 3.02, 3.04, 3.14, 3.32, 3.52, 3.82, 3.87

Using `=QUARTILE.EXC(array, 1)`:
- **Q1 (First Quartile)** = 2.4750

#### (b) Third Quartile (Q3)
Similarly, we use Excel's `=QUARTILE.EXC` function for the third quartile.

Using `=QUARTILE.EXC(array, 3)`:
- **Q3 (Third Quartile)** = 3.3925

#### (c) Mean
To calculate the mean, sum all the GPA values and divide by the number of observations (n = 15).

\[
\text{Mean} = \frac{\sum \text{GPA}}{n} = \frac{2.26 + 3.01 + 3.02 + 2.54 + 3.32 + 3.14 + 3.82 + 2.39 + 3.04 + 2.3 + 2.7 + 3.52 + 2.75 + 2.38 + 3.87}{15} = \frac{43.06}{15} = 2.8707
\]

#### (d) Median
The median is the middle value in the ordered list.

With 15 data points:
- **Median** = 3.01 (the 8th value in the ordered list)

#### (e) Range
Range is the difference between the maximum and minimum values.

\[
\text{Range} = 3.87 - 2.26 = 1.61
\]

#### (f) Standard Deviation
The standard deviation measures the amount of variation or dispersion of a set of values. The formula for sample standard deviation \( s \) is:

\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\]

Calculating it:
- **Standard Deviation** \( s \approx 0.5209 \)

#### (g) Interquartile Range (IQR)
IQR is the difference between the third quartile and the first quartile.

\[
\text{IQR} = Q3 - Q1 = 3.3925 - 2.4750 = 0.9175
\]

#### (h) Lower and Upper Limits for Outliers
To find the limits for outliers, use the following formulas:

\[
\text{Lower Limit (LL)} = Q1 - 1.5 \times \text{IQR}
\]
\[
\text{Upper Limit (UL)} = Q3 + 1.5 \times \text{IQR}
\]

Calculating them:
\[
\text{LL} = 2.4750 - 1.5 \times 0.9175 = 1.0988
\]
\[
\text{UL} = 3.3925 + 1.5 \times 0.9175 = 4.7688
\]

### Summary of Results:
- **(a) Q1** = 2.4750
- **(b) Q3** = 3.3925
- **(c) Mean** = 2.8707
- **(d) Median** = 3.01
- **(e) Range** = 1.61
- **(f) Standard Deviation** = 0.5209
- **(g) IQR** = 0.9175
- **(h) LL** = 1.0988, **UL** = 4.7688

Would you like more details on any of these calculations or explanations?

### 5 Relative Questions:
1. How do you calculate the variance from the standard deviation?
2. What is the significance of the interquartile range in statistics?
3. How does the range compare to the standard deviation in terms of variability?
4. What does it mean if a data point is below the lower limit or above the upper limit for outliers?
5. How would the median change if the data set had an even number of observations?

### Tip:
When calculating quartiles, always ensure your data is sorted in ascending order to avoid errors.

Learn how to analyze GPA statistics including quartiles, mean, median, range, standard deviation, interquartile range (IQR), and outliers. Detailed step-by-step solutions provided.

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