Chapter 11.4, Problem 42HP

To determine

Find the value of $x$ .

Answer to Problem 42HP

  $\boxed{x=7}$

Explanation of Solution

Given information:

A poll of students determined that $88\%$ wanted to go to college. Eight random students are chosen. The probability that at least $x$ students wants to go to college is about $0.752or75.2\%$ . Solve for $x$ .

Calculation:

Consider the survey that finds $88\%$ of students wanted to go to college.

The probability of $k$ successes in $n$ trials of a binomial experiment is,

  $P\left(k\right)=C{}_k{\left(p\right)}^k{\left(q\right)}^{n-k}$

It is given $88\%$ of students wanted to go to college.

Now,

  $\begin{array}{l}p=88\%\\ =0.88\end{array}$

Failure probability is,

  $\begin{array}{l}q=1-p\\ =1-0.88\\ =0.12\end{array}$

The survey was randomly conducted on $8$ students, $k=1,2,\mathrm{3...8}$ in expression $P\left(k\right)=C{}_k{\left(p\right)}^k{\left(q\right)}^{n-k}$ we get,

  $\begin{array}{l}P\left(1\right)=C{}_1{\left(0.88\right)}^1{\left(0.12\right)}^{8-1}\\ =0.0\%\\ P\left(2\right)=C{}_2{\left(0.88\right)}^2{\left(0.12\right)}^{8-2}\\ =0.0\%\\ P\left(3\right)=C{}_3{\left(0.88\right)}^3{\left(0.12\right)}^{8-3}\\ =0.1\%\\ P\left(4\right)=C{}_4{\left(0.88\right)}^4{\left(0.12\right)}^{8-4}\\ =0.9\%\end{array}$

  $\begin{array}{l}P\left(5\right)=C{}_5{\left(0.88\right)}^5{\left(0.12\right)}^{8-5}\\ =5.1\%\\ P\left(6\right)=C{}_6{\left(0.88\right)}^6{\left(0.12\right)}^{8-6}\\ =18.7\%\\ P\left(7\right)=C{}_7{\left(0.88\right)}^7{\left(0.12\right)}^{8-7}\\ =39.2\%\\ P\left(8\right)=C{}_8{\left(0.88\right)}^8{\left(0.12\right)}^{8-8}\\ =36.0\%\end{array}$

Thus $39.2\%+36.0\%=75.2\%$ the statement is valid

Well the probability that $7to8$ wants to go collage is $75.2\%$ .

Hence the value of $\boxed{x=7}$ .