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:

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:


Thus, the sample range is 2.3.
(b)
Compute the sample variance as follows:

![=\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](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B12-1%7D%5Ctimes%20%5B%28180.5-181.41%29%5E%7B2%7D%2B%28181.7-181.41%29%5E%7B2%7D%2B...%2B%28180.5-181.41%29%5E%7B2%7D%5D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B11%7D%5Ctimes%205.7692%5C%5C%5C%5C%3D0.524473%5C%5C%5C%5C%5Capprox%200.5245)
Thus, the sample variance is 0.5245.
(c)
Compute the sample standard deviation as follows:


Thus, the sample standard deviation is 0.7242.
(d)
Compute the sample variance using the shortcut method as follows:
![S^{2}=\frac{1}{n-1}\cdot [\sum x_{i}^{2}-n(\bar x)^{2}]](https://tex.z-dn.net/?f=S%5E%7B2%7D%3D%5Cfrac%7B1%7D%7Bn-1%7D%5Ccdot%20%5B%5Csum%20x_%7Bi%7D%5E%7B2%7D-n%28%5Cbar%20x%29%5E%7B2%7D%5D)
![=\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](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B12-1%7D%5Ccdot%20%5B394913.57-%2812%5Ctimes%20%28181.41%29%5E%7B2%7D%5D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B11%7D%5Ctimes%20%5B394913.57-394907.80%5D%5C%5C%5C%5C%3D%5Cfrac%7B5.77%7D%7B11%7D%5C%5C%5C%5C%3D0.5245)
Thus, the sample variance is 0.5245.