91影视

It is given that

$c=99\%=0.99$

$x=46$

$n=100$

$1-伪=0.99$, using probability table$z_{a/2}=2.575$

The conditions are:

Random: Sample are chosen randomly, so this condition is satisfied.

Independent: Sample of $100$student chips is surely less than $10\%$of all population of role="math" localid="1654452833197" $1800$students.

Normal: Number of success is $46$and failures are $100-46=54$. This is greater than $10$.

All conditions are satisfied.

Sample Proportion: $\hat{p}=\frac{x}{n}$

$=\frac{46}{100}=0.46$

Margin of Error: $E=z_{伪/2}脳\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=2.575脳\sqrt{\frac{0.46(1-0.46)}{100}}=0.1283$

Confidence Interval: $\hat{p}-E=0.46-0.1283=0.3317$

$\hat{p}+E=0.46+0.1283=0.5883$

As total students are $1800$

$1800(0.3317)=597.06鈮�597$

$1800(0.5883)=1058.94鈮�1059$

Hence, number of students that student body present knows lie between$597-1059$