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The percentage of passengers who prefer aisle seats is to be estimated.

The sample size needs to be determined. The following values are given:

The margin of error is equal to 0.025.

The confidence level is equal to 95%.

Let $\hat{p}$denote the sample proportion of passengers who prefer aisle seats.

Let $\hat{q}$denote the sample proportion of passengers who prefer aisle seats.

Here, nothing is known about the percentage to be estimated.

The formula for finding the sample size is as follows:

$n=\frac{{\left[z_{\frac{伪}{2}}\right]}^{2}0.25}{E^2}$

The confidence level is equal to 95%. Thus, the level of significance is equal to 0.05.

The value of $z_{\frac{伪}{2}}$for $伪=0.05$from the standard normal table is equal to 1.96

Substituting the required values, the following value of the sample size is obtained:

$$\begin{gathered} n=\frac{{\left(1.96\right)}^2脳0.25}{{\left(0.025\right)}^2} \\ =1536.64 \\ 鈮�1537 \end{gathered}$$

Hence, the required sample size is 1537.

b.

The value of $\hat{p}$is given to be equal to:

$$\begin{gathered} \hat{p}=38\% \\ =\frac{38}{100} \\ =0.38 \end{gathered}$$

The value of $\hat{q}$is given to be equal to:

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.38 \\ =0.62 \end{gathered}$$

The formula for finding the sample size is as follows:

$n=\frac{{\left[z_{\frac{伪}{2}}\right]}^2\hat{p}\hat{q}}{E^2}$

Substituting the required values, the following value of the sample size is obtained:

$$\begin{gathered} n=\frac{{\left(1.96\right)}^2脳0.38脳0.62}{{\left(0.025\right)}^2} \\ =1448.13 \\ 鈮�1448 \end{gathered}$$

Hence, the required sample size is 1448.