Respuesta :
Answer:
(a) 2.3
(b) 0.5245
(c) 0.7242
(d) 0.5245
Step-by-step explanation:
The data provided is:
S = {180.5, 181.7, 180.9, 181.6, 182.6, 181.6, 181.3, 182.1, 182.1, 180.3, 181.7, 180.5}
(a)
The formula to compute the sample range is:
[tex]\text{Sample Range}=\text{max.}_{x}-\text{min.}_{x}[/tex]
The data set arranged in ascending order is:
S' = {180.3 , 180.5 , 180.5 , 180.9 , 181.3 , 181.6 , 181.6 , 181.7 , 181.7 ,, 182.1 , 182.1 , 182.6}
The minimum value is, 180.3 and the maximum value is, 182.6.
Compute the sample range as follows:
[tex]\text{Sample Range}=\text{max.}_{x}-\text{min.}_{x}[/tex]
[tex]=182.6-180.3\\=2.3[/tex]
Thus, the sample range is 2.3.
(b)
Compute the sample variance as follows:
[tex]S^{2}=\frac{1}{n-1}\sum(x_{i}-\bar x)^{2}[/tex]
[tex]=\frac{1}{12-1}\times [(180.5-181.41)^{2}+(181.7-181.41)^{2}+...+(180.5-181.41)^{2}]\\\\=\frac{1}{11}\times 5.7692\\\\=0.524473\\\\\approx 0.5245[/tex]
Thus, the sample variance is 0.5245.
(c)
Compute the sample standard deviation as follows:
[tex]s=\sqrt{S^{2}}[/tex]
[tex]=\sqrt{0.5245}\\\\=0.7242[/tex]
Thus, the sample standard deviation is 0.7242.
(d)
Compute the sample variance using the shortcut method as follows:
[tex]S^{2}=\frac{1}{n-1}\cdot [\sum x_{i}^{2}-n(\bar x)^{2}][/tex]
[tex]=\frac{1}{12-1}\cdot [394913.57-(12\times (181.41)^{2}]\\\\=\frac{1}{11}\times [394913.57-394907.80]\\\\=\frac{5.77}{11}\\\\=0.5245[/tex]
Thus, the sample variance is 0.5245.
