In a random survey of 226 college students, 20% reported being "only" children (no siblings). The standard deviation is 2%. Determine the confidence interval that is likely to contain the percent of all students nationwide who are only children.

The confidence interval that is likely to contain the percent of all students nationwide who are only children is (0.1505,0.2554)

Step-by-step explanation:

Total number of students = n = 226

We are given that 20% reported being "only" children (no siblings).

Number of students being "only" children (no siblings)

So, x = 46

Z at 95% is 1.96

Formula for confidence interval for sample proportion :

Substitute the values in the formula :

Hence  the confidence interval that is likely to contain the percent of all students nationwide who are only children is (0.1505,0.2554)

g(x) is the negative form of f(x)? because whenever its the negative form it reflects its self over the x axis. idk if this is right though.

step-by-step explanation:

b

step-by-step explanation:

the total is 23.97. 0.25% is off so you have to subtracting 25% from 100% and you will get 75% if you convert it to decimal you will get 0.75. you have to find the original because the formula is % times original = total.so you have the total and the % so you just have to plug it in.