Chapter PIV, Problem 17RE

(a)

To determine

To explain: by guessing with every problem, it is likely to be able to pass.

Explanation of Solution

From the results, note that an airline believes that 5% of passengers do not display flights, overbooks. Suppose there are 265 passengers on a plane, and the airline sells 275 tickets.

Consider X is the random variable representing the number of flight ticket users.

If passengers are alone, then X assumes the binomial distribution with sample size n=275 and the likelihood of success is 0.95 (showing the tickets).

(b)

To determine

To find: 70 percent of the likelihood of having any question correct and how likely it is to pass now.

Explanation of Solution

Here, it is required to determine P (X > 265), since if X is larger than 265 someone (or more than 1) would not get on the plane.

The random variable X is too much large, so it is required to check the normality conditions.

  $\begin{array}{c}np=275\times 0.95\\ =261.25>10\end{array}$

  $\begin{array}{c}nq=n\left(1-p\right)\\ =275\times \left(1-095\right)\\ =275\times 0.05\\ =13.75>10\end{array}$

From the normality conditions, both np and nq have been found to be greater than 10, so use standard binomial approximation to find the necessary probability

(c)

To determine

To find: the probability of the first one getting correct is the 3rd question.

Answer to Problem 17RE

0.1492

Explanation of Solution

Given:

  $\begin{array}{l}n=275\\ p=0.95\end{array}$

Calculation:

The required probability is,

  $\begin{array}{c}P\left(X>265\right)=1-P\left(X\le 265\right)\\ =1-P\left(\frac{X-\mu }{\sigma }\le \frac{256-\mu }{\sigma }\right)\\ =1-P\left(Z\le \frac{256-np}{\sqrt{np\left(1-p\right)}}\right)\\ =1-P\left(Z\le \frac{265-\left(275\times 0.95\right)}{\sqrt{275\times 0.95\times 0.05}}\right)\\ =1-P\left(Z\le 1.04\right)\\ =1-0.8508\\ =0.1492\end{array}$