Answer with explanation:
Given : The age that children learn to walk is normally distributed with [tex]\mu=12\text{ months}[/tex] and [tex]\sigma=1\text{ month}[/tex].
Let x be the age that children learn to walk .
B) Using formula [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
n= 19
For x= 11.5
[tex]z=\dfrac{11.5-12}{\dfrac{1}{\sqrt{19}}}=-2.18[/tex]
For x= 12.5
[tex]z=\dfrac{12.5-12}{\dfrac{1}{\sqrt{19}}}=2.18[/tex]
Then, by using the z-value table , for the 19 people, the probability that the average age that they learned to walk is between 11.5 and 12.5 months old will be :-
[tex]P(-2.18<z<2.18)\\\\=1-2P(z>2.18)\ \ [\because P(-z<Z<z)=1-2P(Z>|z|)\\\\=1-2(1-P(z\leq2.18))\ \ [\because P(Z>z)=1-P(z\leqz) ]\\\\=1-2(1- 0.9853713)\approx0.9707426\approx 0.9707[/tex]
Hence, for the 19 people, the probability that the average age that they learned to walk is between 11.5 and 12.5 months old = 0.9707
C. Using formula [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 11.5
[tex]z=\dfrac{11.5-12}{1}=-0.5[/tex]
For x= 12.5
[tex]z=\dfrac{12.5-12}{1}=0.5[/tex]
Then, by using the z-value table , the probability that the average age that they learned to walk is between 11.5 and 12.5 months old will be :-
[tex]P(-0.5<z<0.5)\\\\=1-2P(z>0.5)\ \ [\because P(-z<Z<z)=1-2P(Z>|z|)\\\\=1-2(1-P(z\leq0.5))\ \ [\because P(Z>z)=1-P(z\leqz) ]\\\\=1-2(1- 0.6914625)\approx0.382925\approx 0.3829[/tex]
Hence, the probability that the average age that they learned to walk is between 11.5 and 12.5 months old = 0.3829
