Answer:
It would be unlikely for them to find fewer than 25 children with the gene.
Step-by-step explanation:
For large numbers the binomial can be approximated closely by a normal distribution.
So mean and standard deviation of the sample according to binomial distribution will be,
[tex]\mu=n\cdot p=950\times 0.06=57[/tex]
[tex]\sigma=\sqrt{n\cdot p\cdot (1-p)}=\sqrt{950\times 0.06\times (1-0.6)}=\sqrt{950\times 0.06\times 0.94}=7.32[/tex]
In normal distribution, we know that
[tex]Z=\dfrac{X-\mu}{\sigma}=\dfrac{25-57}{7.32}=-4.37[/tex]
So the confidence interval would be,
[tex]=\overline X\pm Z\dfrac{s}{\sqrt{n}}[/tex]
Putting the values,
[tex]=57\pm(-4.37)\dfrac{7.32}{\sqrt{950}}\\\\=55.96,58.03[/tex]
As 25 is not in this confidence interval, so it would be unlikely for them to find fewer than 25 children with the gene.
