gpa\n1.54\n2.17\n3.81\n2.30\n3.48\n1.61\n2.16\n2.18\n3.34\n1.9\n2.67\n1.58\n3.79\n3.16\n2.4\n(a) find the…

gpa\n1.54\n2.17\n3.81\n2.30\n3.48\n1.61\n2.16\n2.18\n3.34\n1.9\n2.67\n1.58\n3.79\n3.16\n2.4\n(a) find the first quartile.\n(b) find the third quartile.\n(c) find the mean.\n(d) find the median.\n(e) find the range.\n(f) find s.\n(g) find the inter - quartile range.\n(h) what are the lower and upper limits to find outliers?\n(i) find s².
Answer
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)