91影视

$p=13\%=0.13$

$n=1200$

The mean of a binomial random variable is the product of the sample size $n$and the probability $p$:

localid="1650079570201" $$\begin{gathered} 渭=np \\ =1200(0.13) \\ =156 \end{gathered}$$

The square root of the product of the sample size $n$and the standard deviation of a binomial random variable, the probability $p$and the probability $1-p$:

localid="1650079582715" $$\begin{gathered} 蟽=\sqrt{np(1-p)} \\ =\sqrt{1200(0.13)(1-0.13)} \\ 鈮�11.6499 \end{gathered}$$

Given: the sample contains Hispanic.

$p=13\%=0.13$

We want to know the probability of the number of successes in the sample. Therefore, the distribution is binomial is:

$P(X=k)=\left(\begin{array}{l}n\\ k\end{array}\right)路p^k路(1-p{)}^{n-k}$

We should determine the probability of obtaining $1200脳15\%==180$Hispanics or more extreme:

localid="1650079631185" $$\begin{gathered} P(X鈮�180)=\underset{k=180}{\overset{1200}{鈭�}}\left(\begin{array}{c}n\\ 1200\end{array}\right)路0.{13}^k路(1-0.13{)}^{1200-k} \\ =0.0235 \end{gathered}$$

Since this probability is less than $0.05$or $5\%$, it is unlikely to obtain $180$Hispanics by chance and thus we should be suspicious.