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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:

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

$n(1鈭�p)=1500(1鈭�0.12)=1320>10$

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 铿乶d 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" $$\begin{gathered} P(165鈮�X鈮�195)=P(\frac{165鈭�0.12\left(1500\right)}{\sqrt{1500\left(0.12\right)(1鈭�0.12)}}鈮�Z鈮�\frac{195鈭�0.12\left(1500\right)}{\sqrt{1500\left(0.12\right)(1鈭�0.12)}}) \\ =P(鈭�1.19鈮�Z鈮�1.19) \\ =0.7660 \end{gathered}$$