Chapter 7.2, Problem 31E

(a)

To determine

Whether the 10% condition is met or not.

Answer to Problem 31E

The 10% condition is not met.

Explanation of Solution

Given information:

Number of passengers in the flight = 76

Number of passengers in the first class = 12

Number of passengers in the coach class = 64

Number of passengers for SRS = 10

Given the population size is 76.

Here, the sample size of 10 is more 10% of the population size 76.

Hence, the 10% condition is not met.

(b)

To determine

Whether the Normal condition is met or not.

Answer to Problem 31E

The Normal condition has not been met.

Explanation of Solution

Given information:

Number of passengers in the flight = 76

Number of passengers in the first class = 12

Number of passengers in the coach class = 64

Number of passengers for SRS = 10

Formula used:

When the product of sample size and the sampling proportion that is, $np$ and $n\left(1-p\right)$ both are less than at least 10 then the sampling distribution is approximately Normal.

Consider that the sample distribution be p and sample size for SRS be n.

  $n=76$ and $x=12$

Find the population proportion by dividing the number successes by the sample size that is, $p=\frac{x}{n}$

Substitute 12 for x and 76 for n in the above expression.

  $p=\frac{12}{76}$

Know that the product of sample size and the sampling proportion that is, $np$ and $n\left(1-p\right)$ both are less than at least 10 then the sampling distribution is approximately Normal.

Substitute $\frac{12}{76}$ for p and 10 for n in the expression $np$ .

  $\begin{array}{c}10\times \frac{12}{76}=10\times \frac{3}{19}\\ \approx 1.58\end{array}$

Again, substitute $\frac{12}{76}$ for p and 10 for n in the expression $n\left(1-p\right)$ .

  $\begin{array}{c}10\left(1-\frac{12}{76}\right)=10\times \left(\frac{16}{19}\right)\\ \approx 8.42\end{array}$

It is seen that both $np$ and $n\left(1-p\right)$ are less than 10; so the condition for Normal distribution has not been met.

Hence, the Normal condition has not been met.