Question

2.86
1.88
1.84
2.19
3.11
1.75
2.16
3.81
2.62
3.1
3.43
2.74
(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 interquartile range.
(h) what are the lower and upper limits to find outliers?
(i) find s².

Explanation:

Step1: Calculate the sum of data

Sum = \(2.86 + 1.88+1.84 + 2.19+3.11+1.75+2.16+3.81+2.62+3.1+3.43+2.74\)
Sum = \(31.61\)

Step2: Calculate the mean

There are \(n = 12\) data - points. Mean \(\bar{x}=\frac{\text{Sum}}{n}=\frac{31.61}{12}\approx2.634\)

Step3: Calculate the sample - variance \(s^{2}\)

The formula for the sample - variance is \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
First, calculate \((x_{i}-\bar{x})^{2}\) for each \(x_{i}\):
\((2.86 - 2.634)^{2}=(0.226)^{2}=0.051076\)
\((1.88 - 2.634)^{2}=(- 0.754)^{2}=0.568516\)
\((1.84 - 2.634)^{2}=(-0.794)^{2}=0.630436\)
\((2.19 - 2.634)^{2}=(-0.444)^{2}=0.197136\)
\((3.11 - 2.634)^{2}=(0.476)^{2}=0.226576\)
\((1.75 - 2.634)^{2}=(-0.884)^{2}=0.781456\)
\((2.16 - 2.634)^{2}=(-0.474)^{2}=0.224676\)
\((3.81 - 2.634)^{2}=(1.176)^{2}=1.382976\)
\((2.62 - 2.634)^{2}=(-0.014)^{2}=0.000196\)
\((3.1 - 2.634)^{2}=(0.466)^{2}=0.217156\)
\((3.43 - 2.634)^{2}=(0.796)^{2}=0.633616\)
\((2.74 - 2.634)^{2}=(0.106)^{2}=0.011236\)

\(\sum_{i = 1}^{12}(x_{i}-\bar{x})^{2}=0.051076 + 0.568516+0.630436+0.197136+0.226576+0.781456+0.224676+1.382976+0.000196+0.217156+0.633616+0.011236 = 4.92408\)

\(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}=\frac{4.92408}{11}\approx0.447644\)

Answer:

(c) \(2.634\)
(i) \(0.447644\)