Question

How many passengers are expected to travel between $401$ and $500$ miles and purchase first-class tickets?

Short answer

First-class tickets will cost around $6.6$for customers traveling between $401$ and $500$ miles.

Step-by-step solution

Given Information

Given data:

$\begin{array}{|llllll|}\hline & \text{A}& \text{B}& \text{C}& \text{D}& \text{E}\\ 1& \text{Given Table (Observed Frequency)}& & & & \\ 2& \text{Traveling Distance}& \text{Third Class}& \text{Second Class}& \text{First Class}& \text{Total}\\ 3& 1-100\text{miles}& 21& 14& 6& 41\\ 4& 101-200\text{miles}& 18& 16& 8& 42\\ 5& 201-300\text{miles}& 16& 17& 15& 48\\ 6& 301-400\text{miles}& 12& 14& 21& 47\\ 7& 401-500\text{miles}& 6& 6& 10& 22\\ 8& \text{Total}& 73& 67& 60& 200\\ \hline\end{array}$

Explanation

To calculate the expected frequency is $E=\frac{\text{(rowtotal)(columntotal)}}{\text{(Grandtotal)}}$.

$\begin{array}{|lllll|}\hline & \text{A}& \text{B}& \text{C}& D\\ & \text{Expected Frequency}& & & \\ 1& \text{Traveling Distance}& \text{Third Class}& \text{Second Class}& \text{First Class}\\ 2& 1-100\text{miles}& \frac{\left(73\right)\left(41\right)}{\left(200\right)}=14.965& \frac{\left(67\right)\left(41\right)}{\left(200\right)}=13.735& \frac{\left(60\right)\left(41\right)}{\left(200\right)}=12.3\\ 3& 101-200\text{miles}& \frac{\left(73\right)\left(42\right)}{\left(200\right)}=15.33& \frac{\left(67\right)\left(42\right)}{\left(200\right)}=14.07& \frac{\left(60\right)\left(42\right)}{\left(200\right)}=12.6\\ 4& 201-300\text{miles}& \frac{\left(73\right)\left(48\right)}{\left(200\right)}=17.52& \frac{\left(67\right)\left(48\right)}{\left(200\right)}=16.08& \frac{\left(60\right)\left(48\right)}{\left(200\right)}=14.4\\ 5& 301-400\text{miles}& \frac{\left(73\right)\left(47\right)}{\left(200\right)}=17.155& \frac{\left(67\right)\left(47\right)}{\left(200\right)}=15.745& \frac{\left(60\right)\left(47\right)}{\left(200\right)}=14.1\\ 6& 401-500\text{miles}& \frac{\left(73\right)\left(22\right)}{\left(200\right)}=8.03& \frac{\left(67\right)\left(22\right)}{\left(200\right)}=7.37& \frac{\left(60\right)\left(22\right)}{\left(200\right)}=6.6\\ \hline\end{array}$

First-class tickets will cost around $6.6$ for customers traveling between $401$ and $500$ miles.