Question

1. the data in the table below shows the number of passengers and number of suitcases on various airplanes.

passengers	suitcases
92	180
115	239
128	272
143	290
154	310
178	357
200	405

a) find the line of best fit:

b) estimate the number of suitcases on a flight carrying 250 people.

Explanation:

Step 1: Calculate necessary sums

Let \( x \) be the number of passengers and \( y \) be the number of suitcases.
We have \( n = 8 \) data points.
First, calculate \( \sum x \), \( \sum y \), \( \sum xy \), and \( \sum x^2 \):

• \( \sum x = 75 + 92 + 115 + 128 + 143 + 154 + 178 + 200 = 1085 \)
• \( \sum y = 159 + 180 + 239 + 272 + 290 + 310 + 357 + 405 = 2212 \)
• \( \sum xy = (75×159)+(92×180)+(115×239)+(128×272)+(143×290)+(154×310)+(178×357)+(200×405) \)

\( = 11925 + 16560 + 27485 + 34816 + 41470 + 47740 + 63546 + 81000 = 324542 \)

• \( \sum x^2 = 75^2 + 92^2 + 115^2 + 128^2 + 143^2 + 154^2 + 178^2 + 200^2 \)

\( = 5625 + 8464 + 13225 + 16384 + 20449 + 23716 + 31684 + 40000 = 159547 \)

Step 2: Calculate slope \( m \) and intercept \( b \)

The formula for the slope \( m \) of the line of best fit is:
\( m=\frac{n\sum xy - \sum x\sum y}{n\sum x^2 - (\sum x)^2} \)
Substitute the values:
\( m=\frac{8×324542 - 1085×2212}{8×159547 - 1085^2} \)
First, calculate numerator: \( 8×324542 = 2596336 \), \( 1085×2212 = 1085×2200 + 1085×12 = 2387000 + 13020 = 2400020 \)
Numerator: \( 2596336 - 2400020 = 196316 \)
Denominator: \( 8×159547 = 1276376 \), \( 1085^2 = 1177225 \)
Denominator: \( 1276376 - 1177225 = 99151 \)
So, \( m=\frac{196316}{99151} \approx 1.98 \) (approx 2)

The formula for the intercept \( b \) is:
\( b=\frac{\sum y - m\sum x}{n} \)
Substitute \( m \approx 2 \), \( \sum y = 2212 \), \( \sum x = 1085 \), \( n = 8 \):
\( b=\frac{2212 - 2×1085}{8}=\frac{2212 - 2170}{8}=\frac{42}{8}=5.25 \)

So the line of best fit is \( y = 2x + 5.25 \) (or more accurately with \( m \approx 1.98 \), \( y \approx 1.98x + 5.25 \))

Step 3: Estimate for 250 passengers (part b)

Use the line of best fit \( y = 2x + 5.25 \) (using the approximate slope for simplicity). Substitute \( x = 250 \):
\( y = 2×250 + 5.25 = 500 + 5.25 = 505.25 \)
Using the more accurate \( m \approx 1.98 \):
\( y = 1.98×250 + 5.25 = 495 + 5.25 = 500.25 \approx 500 \) (or 505 depending on the slope used)

Answer:

(for part a):
The line of best fit is approximately \( \boldsymbol{y = 2x + 5.25} \) (or more accurately \( y \approx 1.98x + 5.25 \))