A survey showed that 81% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 11 adults are randomly selected, find the probability that no more than 1 of them need correction for their eyesight. Is 1 a significantly low number of adults requiring eyesight correction?

The probability that no more than 1 of the 11 adults require eyesight correction is

5.579 * 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 *

2. 1 adult need correction = 1C11 * *

Then the probability will be:

0C11 * +1C11 * * =

 *     +     * *0.81 =

5.579 * = 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?

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all

its a i did it myself

step-by-step explanation: