Data handling

Advanced Math 7 · Unit 15 · data-handling-7 · Teacher edition

data-handling-7

Name ______________________________Date ______________

1.
Which of these is continuous data?
1. (A)
  The heights of students, in cm
2. (B)
  The number of students in each class
3. (C)
  The shoe sizes sold in a shop
4. (D)
  The number of goals scored in each match
2.
A back-to-back stem-and-leaf plot shows test scores for two classes. Leaves are read outward from the stem, so the top row means Class A scored $58$ and $55$ , and Class B scored $52$ and $56$ .

Class A leaves Stem Class B leaves
8 5 5 2 6
7 3 1 6 0 4 4 8
6 2 7 1 5 9
0 8 3

Find the median of Class B minus the median of Class A.
Answer: ______________
3.
Eight students recorded their hours of sleep on Saturday night: $6, 7, 7, 7, 8, 8, 9, 20$ . Find the mean minus the median.
Answer: ______________
4.
The table shows how long $20$ students spent on homework.

Minutes 0–10 10–20 20–30
Students $4$ $10$ $6$

Estimate the mean time, in minutes.
1. (A)
  $10$
2. (B)
  $15$
3. (C)
  $16$
4. (D)
  $20$
5.
The mean of five numbers is $12$ . When one of the numbers is removed, the mean of the remaining four is $10$ . What number was removed?
1. (A)
  $2$
2. (B)
  $10$
3. (C)
  $12$
4. (D)
  $20$
5. (E)
  $22$

Minutes	0–10	10–20	20–30
Students	$4$	$10$	$6$

Answer key — Data handling

1.
(A)
The heights of students, in cm
data-handling-7-01
Heights are measured and can take any value in a range, so they are continuous. Numbers of students and goals are counts; shoe sizes come in fixed steps — all discrete.
2.
1data-handling-7-02
Class A median $= \frac{63 + 67}{2} = 65$ . Class B: $52, 56, 60, 64, 64, 68, 71, 75, 79, 83$ , median $= \frac{64 + 68}{2} = 66$ . Difference $= 66 - 65 = \mathbf{1}$ . (Class B is also more spread out: range $31$ versus $25$ .)
3.
1.5data-handling-7-03
Mean $= \frac{72}{8} = 9$ ; median $= \frac{7 + 8}{2} = 7.5$ . Difference $= \mathbf{1.5}$ . The single $20$ pulls the mean above every other value; the median, $7.5$ , is the better “typical” night.
4.
(C)
$16$
data-handling-7-04
$\frac{5(4) + 15(10) + 25(6)}{20} = \frac{320}{20} = \mathbf{16}$ minutes. ( $10$ uses the interval width $10$ for every student; $15$ is just the middle interval's midpoint; $20$ is the number of students.)
5.
(D)
$20$
data-handling-7-05
Total before $= 60$ ; total after $= 40$ . Removed number $= 60 - 40 = \mathbf{20}$ . ( $2$ is just $12 - 10$ ; $22$ adds the change instead of subtracting it.)