Question

the weights in pounds of some male and female dogs at an animal shelter are listed below.
male: 6, 12, 15, 25, 46, 75, 10, 82, 103
female: 4, 11, 14, 20, 44, 65, 8, 76, 96
which statement is true?
the standard deviation of male dogs is greater than the standard deviation of female dogs by approximately 4.2.
the standard deviation of female dogs is greater than the standard deviation of male dogs by approximately 2.5.
the standard deviation of male dogs is greater than the standard deviation of female dogs by approximately 2.5.
the standard deviation of female dogs is greater than the standard deviation of male dogs by approximately 4.2.

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.