He accompanying data table describes results from groups of 10 births from 10 different sets of parents. the random variable x represents the number of girls among 10 children. complete the questions below. loading... click the icon to view the data table.

Given the following data table that describes results from groups of 10 births from 10 different sets of parents. The random variable x represents the number of girls among 10 children.

[tex]\begin{tabular} {|c|c|} x&p(x)\\[1ex] 0&0.005\\ 1&0.014\\ 2&0.037\\ 3&0.114\\ 4&0.208\\ 5&0.241\\ 6&0.202\\ 7&0.116\\ 8&0.037\\ 9&0.013\\ 10&0.013\\[1ex] &\Sigma p(x)=1 \end{tabular}[/tex]The range rule of thumb says that the range is about four times the standard deviation. We obtain the standard deviation as follows:[tex]\begin{tabular} {|c|c|p{2.3cm}|c|c|} x&p(x)&xp(x)&\left(x-E(x)\right)&\left(x-E(x)\right)^2\\[1ex] 0&0.005&0&-5.034&25.341\\ 1&0.014&0.014&-4.034&16.273\\ 2&0.037&0.074&-3.034&9.205\\ 3&0.114&0.342&-2.034&4.137\\ 4&0.208&0.832&-1.034&1.069\\ 5&0.241&1.205&-0.034&0.001\\ 6&0.202&1.212&0.966&0.933\\ 7&0.116&0.812&1.966&3.865\\ 8&0.037&0.296&2.966&8.797\\ 9&0.013&0.117&3.966&15.729\\ 10&0.013&0.13&4.966&24.661\\[1ex] &\Sigma p(x)=1&E(x)=\Sigma xp(x)=5.034 \end{tabular}[/tex][tex]\begin{tabular} {|c|} \left(x-E(x)\right)^2p(x)\\[1ex] 0.127\\0.228\\0.341\\0.472\\0.222\\0.000\\0.188\\0.448\\0.325\\0.204\\0.321\\[1ex]\Sigma\left(x-E(x)\right)^2p(x)=2.876 \end{tabular}[/tex]The standard deviation is given by[tex] \sqrt{\left(x-E(x)\right)^2p(x)} = \sqrt{2.876} =1.696[/tex]Therefore, the range of values containing the usual numbers of girls in 10 births is given by:[tex]E(x)\pm2\sigma=5.034\pm2(1.696) \\ \\ =5.034\pm3.392=(5.034-3.392,\ 5.034+3.392) \\ \\ =(1.642,\ 8.426)[/tex] Based on the​ result, 1 girl in 10 births is an unusually low number of​ girls because 1 is not within the normal range of values of number of girls per 10 births.