On the Web, What kinds of Web sites do males aged $18$to $34$visit most often? Half of the male Internet users in this age group visit an auction site such as eBay at least once a month.$11$A study of Internet use interviews a random sample of $500$men aged $18$to $34$. Let $X$= the number in the sample who visit an auction site at least once a month.

(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 at least $235$of the men in the sample visit an online auction site at least once a month.

It is given in the question that, the probability of success (p)=$0.50$

Number of trials(n) = $500$

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

The random variable $X$is a binomial random variable, which means it has two sides:

• There are two types of successes and failures: going to auction and not going to auction.
• Males are autonomous from one another.
• The number of males is predetermined.
• The probability of selecting a specific person is fixed.

It is given in the question that, the probability of success $\left(p\right)=0.50$

A number of trials$\left(n\right)=500$

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

The following are the conditions for normal approximation:

$$\begin{gathered} np=500\left(0.5\right) \\ =250>10 \end{gathered}$$

$$\begin{gathered} n(1-p)=500(1-0.5) \\ =250>10 \end{gathered}$$

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

It is given in the question that, the probability of success $\left(p\right)=0.50$

A number of trials$\left(n\right)=500$

Use the Normal approximation to find the probability that at least $235$of the men in the sample visit an online auction site at least once a month.

The probability of at least $235$men is computed as:

localid="1650032037644" $$\begin{gathered} P(X\ge 235)=P(Z\ge \frac{235-0.5\left(1500\right)}{\sqrt{500\left(0.50\right)(1-0.50)}}) \\ =P(Z\ge -1.34) \\ =0.9099 \end{gathered}$$