Question

gpa
1.65
2.71
3.14
2
2.61
1.59
1.82
2.91
1.66
1.97
2.06
2.5
1.9
2.56
2.28
(a) find the first quartile.
(b) find the third quartile.
(c) find $\bar{x}$.
(d) find the median.
(e) find the range.
(f) find s.

Explanation:

Step1: Sort the data

$1.59, 1.65, 1.66, 1.82, 1.9, 1.97, 2, 2.06, 2.28, 2.5, 2.56, 2.61, 2.71, 2.91, 3.14$

Step2: Calculate the position of the first - quartile

$n = 15$, $Q_1$ position $= \frac{n + 1}{4}=\frac{15+ 1}{4}=4$. So $Q_1$ is the 4th value in the sorted data.
$Q_1 = 1.82$

Step3: Calculate the position of the third - quartile

$Q_3$ position $= \frac{3(n + 1)}{4}=\frac{3\times(15 + 1)}{4}=12$. So $Q_3$ is the 12th value in the sorted data.
$Q_3=2.61$

Step4: Calculate the mean $\bar{x}$

$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{1.59+1.65+1.66+1.82+1.9+1.97+2+2.06+2.28+2.5+2.56+2.61+2.71+2.91+3.14}{15}=\frac{34.36}{15}\approx2.29$

Step5: Calculate the median

Since $n = 15$ (odd), the median is the $\frac{n + 1}{2}$ - th value. $\frac{15+1}{2}=8$ - th value. Median $= 2.06$

Step6: Calculate the range

Range $= \text{Max}-\text{Min}=3.14 - 1.59 = 1.55$

Step7: Calculate the sample standard deviation $s$

First, calculate the sum of squared differences from the mean.
$S=\sum_{i = 1}^{n}(x_i-\bar{x})^2$.
$(1.59 - 2.29)^2+(1.65 - 2.29)^2+(1.66 - 2.29)^2+(1.82 - 2.29)^2+(1.9 - 2.29)^2+(1.97 - 2.29)^2+(2 - 2.29)^2+(2.06 - 2.29)^2+(2.28 - 2.29)^2+(2.5 - 2.29)^2+(2.56 - 2.29)^2+(2.61 - 2.29)^2+(2.71 - 2.29)^2+(2.91 - 2.29)^2+(3.14 - 2.29)^2$
$S=(- 0.7)^2+(-0.64)^2+(-0.63)^2+(-0.47)^2+(-0.39)^2+(-0.32)^2+(-0.29)^2+(-0.23)^2+(-0.01)^2+(0.21)^2+(0.27)^2+(0.32)^2+(0.42)^2+(0.62)^2+(0.85)^2$
$S = 0.49+0.4096+0.3969+0.2209+0.1521+0.1024+0.0841+0.0529+0.0001+0.0441+0.0729+0.1024+0.1764+0.3844+0.7225=3.4116$
$s=\sqrt{\frac{S}{n - 1}}=\sqrt{\frac{3.4116}{14}}\approx0.49$

Answer:

(a) $1.82$
(b) $2.61$
(c) $2.29$
(d) $2.06$
(e) $1.55$
(f) $0.49$