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gpa 1.54 2.17 3.81 2.30 3.48 1.61 2.16 2.18 3.34 1.9 2.67 1.58 3.79 3.1…

Question

gpa
1.54
2.17
3.81
2.30
3.48
1.61
2.16
2.18
3.34
1.9
2.67
1.58
3.79
3.16
2.4
(a) find the first quartile.
(b) find the third quartile.
(c) find the mean.
(d) find the median.
(e) find the range.
(f) find s.
(g) find the inter - quartile range.
(h) what are the lower and upper limits to find outliers?
(i) find s².

Explanation:

Step1: Sort the data

1.54, 1.58, 1.61, 1.9, 2.16, 2.17, 2.18, 2.4, 2.67, 3.16, 3.34, 3.48, 3.79, 3.81
There are \(n = 14\) data - points.

Step2: Calculate the first - quartile \(Q_1\)

The position of \(Q_1\) is \(i=\frac{n + 1}{4}=\frac{14+1}{4}=3.75\).
\(Q_1\) is \(0.25\) of the way between the 3rd and 4th ordered data - points.
The 3rd value is \(1.61\) and the 4th value is \(1.9\).
\(Q_1=1.61+(1.9 - 1.61)\times0.25=1.61 + 0.0725=1.6825\)

Step3: Calculate the third - quartile \(Q_3\)

The position of \(Q_3\) is \(i = \frac{3(n + 1)}{4}=\frac{3\times(14 + 1)}{4}=11.25\).
\(Q_3\) is \(0.25\) of the way between the 11th and 12th ordered data - points.
The 11th value is \(3.34\) and the 12th value is \(3.48\).
\(Q_3=3.34+(3.48 - 3.34)\times0.25=3.34+0.035 = 3.375\)

Step4: Calculate the mean \(\bar{x}\)

\(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{1.54+1.58+1.61+1.9+2.16+2.17+2.18+2.4+2.67+3.16+3.34+3.48+3.79+3.81}{14}\)
\(\sum_{i = 1}^{n}x_i=34.89\)
\(\bar{x}=\frac{34.89}{14}\approx2.4921\)

Step5: Calculate the median

Since \(n = 14\) (an even number), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th ordered data - points.
The 7th value is \(2.18\) and the 8th value is \(2.4\).
Median\(=\frac{2.18 + 2.4}{2}=2.29\)

Step6: Calculate the range

Range\(=\text{Max}-\text{Min}=3.81 - 1.54 = 2.27\)

Step7: Calculate the standard deviation \(s\)

First, calculate the variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}\)
\((x_1-\bar{x})^2=(1.54 - 2.4921)^2\), \((x_2-\bar{x})^2=(1.58 - 2.4921)^2,\cdots,(x_{14}-\bar{x})^2=(3.81 - 2.4921)^2\)
\(\sum_{i = 1}^{n}(x_i-\bar{x})^2 = 10.9977\)
\(s^{2}=\frac{10.9977}{13}\approx0.846\)
\(s=\sqrt{0.846}\approx0.92\)

Step8: Calculate the inter - quartile range (IQR)

\(IQR=Q_3 - Q_1=3.375-1.6825 = 1.6925\)

Step9: Calculate the lower and upper limits for outliers

Lower limit\(=Q_1-1.5\times IQR=1.6825-1.5\times1.6925=1.6825 - 2.53875=- 0.85625\)
Upper limit\(=Q_3 + 1.5\times IQR=3.375+1.5\times1.6925=3.375+2.53875 = 5.91375\)

Step10: Calculate the variance \(s^{2}\) (already calculated in Step 7)

\(s^{2}\approx0.846\)

Answer:

(a) \(1.6825\)
(b) \(3.375\)
(c) \(2.4921\)
(d) \(2.29\)
(e) \(2.27\)
(f) \(0.92\)
(g) \(1.6925\)
(h) \(L=-0.85625\), \(U = 5.91375\)
(i) \(0.846\)