Answer:
The maximum value in the range is 8.113
1 girl in 10 births is a significantly low number of girls.
Explanation:
Note that the range rule of thumb says that the range of about 4 times the standard deviation.
We'll use the below formula to determine the standard deviation;
[tex]\begin{gathered} \sigma=\sqrt[]{\lbrack\sum^{}_{}x^2\cdot P(x)\rbrack-\mu^2} \\ \text{where }\mu=\text{ population mean} \end{gathered}[/tex]Let's go ahead and determine the mean as seen below;
[tex]\mu=\sum ^{}_{}\lbrack x\cdot P(x)\rbrack[/tex][tex]\begin{gathered} \mu=(0\cdot0.005)+(1\cdot0.12)+(2\cdot0.039)+(3\cdot0.113)+(4\cdot0.196)+(5\cdot0.235)+(6\cdot0.209) \\ +(7\cdot0.113)+(8\cdot0.036)+(9\cdot0.016)+(10\cdot0.026) \end{gathered}[/tex][tex]\begin{gathered} \mu=0.12+0.078+0.339+0.784+1.175+1.254+0.791+0.288+0.144+0.26 \\ \mu=5.233 \end{gathered}[/tex]Let's now determine the below;
[tex]\begin{gathered} \sum ^{}_{}x^2\cdot P(x)=(0^2\cdot0.005)+(1^2\cdot0.12)+(2^2\cdot0.039)+(3^2\cdot0.113)+(4^2\cdot0.196)+(5^2\cdot0.235) \\ +(6^2\cdot0.209)+(7^2\cdot0.113)+(8^2\cdot0.036)+(9^2\cdot0.016)+(10^2\cdot0.026) \end{gathered}[/tex][tex]\begin{gathered} \sum ^{}_{}x^2\cdot P(x)=0.012+0.156+1.017+3.136+5.875+7.524+5.537+2.304+1.296+2.6 \\ =29.457 \end{gathered}[/tex]So the standard deviation will be;
[tex]\sigma=\sqrt[]{29.457-5.233^2}=\sqrt[]{29.457-27.384}=\sqrt[]{2.073}=1.44[/tex]Let's determine the maximum and minimum value of the distribution as seen below;
[tex]\begin{gathered} Maximum\text{ value = }\mu+2\sigma=5.233+2(1.44)=5.233+2.88=8.113 \\ \text{Minimum value }=\mu-2\sigma=5.233-2(1.44)=5.233-2.88=2.353 \end{gathered}[/tex]We can see from the above that the number of girls born among 10 children should be between the range of 2.353 and 8.113, therefore 1 girl in 10 births is a significantly low number of girls.
The maximum value in this range is 8.113