Answer:
5.579 * [tex]10^{-7}[/tex] or 0.00005579 %
Step-by-step explanation:
In this question, you select 11 random adults, so the order is not important and we should use combination instead of permutation. There are two different probability here
A= chance that the adults need correction= 0.81
B= chance that the adult doesn't need correction = 0.19
The case that can fulfill the condition of no more than 1 of 11 adults need correction is:
1. 0 adult need correction = 0C11 * [tex]B^{11}[/tex]
2. 1 adult need correction = 1C11 * [tex]A^{1}[/tex] * [tex]B^{10}[/tex]
Then the probability will be:
0C11 * [tex]B^{11}[/tex] +1C11 * [tex]A^{1}[/tex] * [tex]B^{10}[/tex]=
[tex]\frac{11!}{0!(11-0)!}[/tex] * [tex]0.19^{11}[/tex] + [tex]\frac{11!}{1!(11-1)!}[/tex] * [tex]0.19^{10}[/tex] *0.81 =
5.579 * [tex]10^{-7}[/tex]= 0.00005579 %
The threshold for a significant value that widely used is 5% and the chance is lower than 5%, so no more than 1 adult need correction is a significantly low number of adults requiring eyesight correction?