Estimating the Mean for Grouped Data

Worked example

1. 1

   $5\times 4=20$ — Row 1: midpoint × frequency. We count each of the 4 students as taking 5 minutes.

2. 2

   $15\times 14=210$ — Row 2: the 14 students in 10 to 20 minutes each count as 15 minutes.

3. 3

   $25\times 12=300$ — Row 3: the 12 students in 20 to 30 minutes each count as 25 minutes.

4. 4

   $35\times 10=350$ — Row 4: the 10 students in 30 to 40 minutes each count as 35 minutes.

5. 5

   $\sum fx=20+210+300+350=880$ — Add the four products. This is an estimate of all 40 times added together.

6. 6

   $\sum f=4+14+12+10=40$ — Add the frequencies to get how many students there are.

7. 7

   $\frac{880}{40}=22\text{ minutes}$ — Mean = total ÷ how many. 22 lies between 0 and 40, so it is sensible.

Worked example

1. 1

   $\frac{2+5}{2}=3.5$ — Midpoint of 2 to 5: add the ends and halve. A midpoint can be a decimal.

2. 2

   $1,3.5,7.5,15$ — Do the same for every class: (0 + 2) ÷ 2, (5 + 10) ÷ 2, (10 + 20) ÷ 2. Write the midpoints down — they earn a mark.

3. 3

   $1\times 6+3.5\times 14+7.5\times 20+15\times 10$ — Multiply each midpoint by its frequency, row by row.

4. 4

   $=6+49+150+150=355$ — Work out each product, then add them. This is $\sum fx$.

5. 5

   $6+14+20+10=50$ — Add the frequencies: 50 parcels, as the question said.

6. 6

   $\frac{355}{50}=7.1\text{ kg}$ — Divide. 7.1 kg lies between 0 and 20 kg, so it is sensible.

Worked example

1. 1

   $10\times 4=40$ — The $0 to $20 class has midpoint 10. Count each of its 4 weeks as $10.

2. 2

   $30\times 10=300$ — The $20 to $40 class has midpoint 30, and it happened in 10 weeks.

3. 3

   $50\times 6=300$ — The $40 to $60 class has midpoint 50, and it happened in 6 weeks.

4. 4

   $40+300+300=640$ — Add the products. This is the estimated total, $640. Stop here: there is no dividing.

Worked example

1. 1

   $5\times 6+25\times 9+35\times 5=430$ — Midpoint × frequency for the three rows you know: 30 + 225 + 175.

2. 2

   $\sum fx=430+15x$ — The missing row has midpoint 15 and frequency x, so it adds 15 × x = 15x.

3. 3

   $\sum f=20+x$ — The known frequencies make 6 + 9 + 5 = 20. Add x too: those people count.

4. 4

   $\frac{430+15x}{20+x}=20$ — Mean = Σfx ÷ Σf, and the question says the mean is 20.

5. 5

   $430+15x=20(20+x)$ — Multiply both sides by (20 + x) to clear the fraction.

6. 6

   $430+15x=400+20x$ — Expand the bracket: 20 × 20 = 400 and 20 × x = 20x.

7. 7

   $30=5x$ — Take 400 and 15x from both sides, so x is on one side only.

8. 8

   $x=6$ — Divide both sides by 5.

9. 9

   $\frac{430+15\times 6}{20+6}=\frac{520}{26}=20$ — Check: put x = 6 back in. We get the given mean, 20, so x = 6 is right.

Worked example

1. 1

   $\frac{40+1}{2}=20.5$ — There are 40 students, so the middle one is at position 20.5, halfway between the 20th and 21st.

2. 2

   $4$ — Running total after the first class, 0 to 10: 4 students. Not yet 20.5.

3. 3

   $4+14=18$ — Add the next class, 10 to 20. 18 students so far, still short of 20.5.

4. 4

   $18+12=30$ — Add the class 20 to 30. Now 30 students, which passes 20.5.

5. 5

   $\text{median class: }20<t\le 30$ — The running total passed the middle inside this class, so the median is in it.