the weights in pounds of some male and female dogs at an animal shelter are listed below.\nmale: 6, 12, 15…

Explanation:

Step1: Calculate mean of male dogs

Let (x_{male}={6, 12, 15, 25, 46, 75, 10, 82, 103}), (\bar{x}_{male}=\frac{6 + 12+15+25+46+75+10+82+103}{9}=\frac{374}{9}\approx41.56)

Step2: Calculate variance of male dogs

(s_{male}^2=\frac{\sum_{i = 1}^{9}(x_{i}-\bar{x}{male})^2}{9 - 1}) ((6 - 41.56)^2+(12 - 41.56)^2+(15 - 41.56)^2+(25 - 41.56)^2+(46 - 41.56)^2+(75 - 41.56)^2+(10 - 41.56)^2+(82 - 41.56)^2+(103 - 41.56)^2) (=(- 35.56)^2+(-29.56)^2+(-26.56)^2+(-16.56)^2+(4.44)^2+(33.44)^2+(-31.56)^2+(40.44)^2+(61.44)^2) (=1264.51+873.89+705.43+274.23+19.71+1118.23+995.03+1635.39+3774.87) (=\frac{10661.3}{8}\approx1332.66) (s{male}=\sqrt{1332.66}\approx36.5)

Step3: Calculate mean of female dogs

Let (x_{female}={4, 11, 14, 20, 44, 65, 8, 76, 96}), (\bar{x}_{female}=\frac{4+11+14+20+44+65+8+76+96}{9}=\frac{338}{9}\approx37.56)

Step4: Calculate variance of female dogs

(s_{female}^2=\frac{\sum_{i = 1}^{9}(x_{i}-\bar{x}{female})^2}{9 - 1}) ((4 - 37.56)^2+(11 - 37.56)^2+(14 - 37.56)^2+(20 - 37.56)^2+(44 - 37.56)^2+(65 - 37.56)^2+(8 - 37.56)^2+(76 - 37.56)^2+(96 - 37.56)^2) (=(-33.56)^2+(-26.56)^2+(-23.56)^2+(-17.56)^2+(6.44)^2+(27.44)^2+(-29.56)^2+(38.44)^2+(58.44)^2) (=1126.27+705.43+555.07+308.35+41.47+752.95+873.89+1477.63+3415.23) (=\frac{9256.2}{8}\approx1157.03) (s{female}=\sqrt{1157.03}\approx34.0)

Step5: Compare standard - deviations

(s_{male}-s_{female}=36.5 - 34.0 = 2.5)

Answer:

The standard deviation of male dogs is greater than the standard deviation of female dogs by approximately 2.5.