Checking for survey errors One way of checking the effect of undercoverage, nonresponse, and other sources of error in a sample survey is to compare the sample with known facts about the population. About $12$% of American adults identify themselves as black. Suppose we take an SRS of $1500$American adults and let $X$be the number of blacks in the sample.

(a) Show that $X$is approximately a binomial random variable.

(b) Check the conditions for using a Normal approximation in this setting.

(c) Use the Normal approximation to find the probability that the sample will contain between $165$and $195$blacks. Show your work.

It is given in the question that, the probability of success(p)=$12\%$$=0.12$

Number of trials (n)=$1500$

Show that $X$is approximately a binomial random variable.

The random variable $X$is a binomial random variable, which means that it has the following properties:

• There are two types of success and failures: black and non-black.
• The United States of America is self-sufficient.
• The amount of blacks is predetermined.
• The likelihood of selecting a black person is fixed.

Because all of the requirements for a binomial distribution are predetermined. As a result, the variable $X$is a random variable with a binomial distribution.

It is given in the question that, the probability of success$\left(p\right)=12\%$=$0.12$

Number of trialslocalid="1649681024831" $\left(n\right)=1500$

Check the conditions for using a normal approximation in this setting

Following are the condition for normal approximation is:

$np=1500\left(0.12\right)=180>10$

$n(1-p)=1500(1-0.12)=1320>10$

As a result, a normal binomial distribution approximation might be utilized.

It is given in the question that, the probability of success(p)= $12\%$= $0.12$

Number of trials (n)= $1500$

Use the Normal approximation to find the probability that the sample will contain between $165$and $195$ blacks.

The probability that the sample would contain $165$and $195$blacks is computed as

localid="1650032315279" $$\begin{gathered} P(165\le X\le 195)=P(\frac{165-0.12\left(1500\right)}{\sqrt{1500\left(0.12\right)(1-0.12)}}\le Z\le \frac{195-0.12\left(1500\right)}{\sqrt{1500\left(0.12\right)(1-0.12)}}) \\ =P(-1.19\le Z\le 1.19) \\ =0.7660 \end{gathered}$$