Chapter 19, Problem 19.63SE

a.

To determine

Theprobability that exactly eight of the 12 visit an auction site in past month.

Answer to Problem 19.63SE

The approximate probability that exactly 8 out of 12 students visit an auction site is

0.1208.

Explanation of Solution

Given info:

In the age group 18 to 34, 50% of male internet users visit an auction site at least once a month. An interview done with 12 male internet users, out of those exactly eight visit the auction site in past month.

Calculation:

Convert the probability into decimal form,

$\begin{array}{c}p=\frac{50}{100}\\ =0.5\end{array}$

The formula to calculate probability that exactly 8 out of 12 students visit an auction site is,

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

Where n is number of units.

Simplify the above formula.

$P\left(X=k\right)=\left(\frac{n!}{k!\left(n-k\right)!}\right)p^k{\left(1-p\right)}^{n-k}\left[\because \left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n!}{k!\left(n-k\right)!}\right]$

Substitute 8 for k, 12 for n and 0.5 for p in above formula.

$\begin{array}{c}P\left(X=8\right)=\left(\frac{12!}{8!\left(12-8\right)!}\right){\left(0.5\right)}^8{\left(1-0.5\right)}^{12-8}\\ =\left(\frac{12!}{8!4!}\right)\left(0.00390625\right){\left(0.5\right)}^4\\ =\left(\frac{9\times 10\times 11\times 12}{1\times 2\times 3\times 4}\right)\left(0.00390625\right)\left(0.0625\right)\end{array}$

Further simplify to find probability.

$\begin{array}{c}P\left(X=8\right)=495\left(2.4413\times {10}^{-4}\right)\\ =0.1208\end{array}$

Thus, the approximate probability that exactly 8 out of 12 students visit an auction site is

0.1208.

b.

To determine

To find: The probability that at least 235 men in the sample visit online auction site at least once a month.

Answer to Problem 19.63SE

The probability that at least 235 men in the sample visit online auction site at least once a month is 0.9099.

Explanation of Solution

Given info:

In the age group 18 to 34, 50% of male internet users visit an auction site at least once a month. An interview done with 500 male internet users aged 18 to 34.

Calculation:

Convert the probability into decimal form,

$\begin{array}{c}p=\frac{50}{100}\\ =0.5\end{array}$

Use normal approximation equation to check that normal approximation is appropriate.

$np\ge 10$

Substitute 500 for n and 0.5 for p in above formula.

$\begin{array}{c}500\times 0.5\ge 10\\ 250\ge 10\end{array}$

So, the normal approximation is correct.

Suppose, $X\sim \text{B}\left(n,p\right)$ and if n is large or p is 0.5, then $X\sim \text{N}\left(np,npq\right)$.

The required probability is calculated as:

$\begin{array}{c}P\left(X>235\right)=P\left(\frac{X-np}{\sqrt{npq}}>\frac{235-np}{\sqrt{npq}}\right)\left(n=500,p=0.5,q=0.5\right)\\ =P\left(\frac{X-500\times 0.5}{\sqrt{500\times 0.5\times 0.5}}>\frac{235-500\times 0.5}{\sqrt{500\times 0.5\times 0.5}}\right)\\ =P\left(Z>\frac{-15}{11.180}\right)\\ =P\left(Z>-1.341\right)\end{array}$

Further solve the above expression.

$\begin{array}{c}P\left(X>235\right)=P\left(Z<1.341\right)\left(\text{By}\text{symmetric}\right)\\ =0.9099\left(Z\text{-table}\right)\end{array}$

Thus, the probability that at least 235 men in the sample visit online auction site at least once a month is 0.9099.