Percentiles Quartiles Scaffolded Xabdw2 | PDF | Descriptive ... - Free Printable

Let’s solve each problem step by step.

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Problem 1: Find the range for each set of data.

Range = largest value – smallest value

a) Data: 6, 8, 11, 15, 24, 38
Largest = 38, Smallest = 6
Range = 38 - 6 = 32

b) Data: 11, -6, -2, 16, 9, -8, 17, 19
Largest = 19, Smallest = -8
Range = 19 - (-8) = 19 + 8 = 27

c) Data: 6.4, 3.8, 5.9, 4.7, 5.3, 7.1, 3.2
Largest = 7.1, Smallest = 3.2
Range = 7.1 - 3.2 = 3.9

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Problem 2: For the data below, find median, Q3, and Q1.

Data: 6, 47, 49, 15, 43, 41, 7, 39, 43, 41, 36

Step 1: Order the data from smallest to largest.

Ordered data:
6, 7, 15, 36, 39, 41, 41, 43, 43, 47, 49

There are 11 numbers (odd count).

Median is the middle number → 6th number in ordered list → 41

Now split into lower half and upper half (exclude median since odd count):

Lower half (first 5 numbers): 6, 7, 15, 36, 39 → Median of this = Q1 = 15

Upper half (last 5 numbers): 43, 43, 47, 49 → Wait, that’s only 4? Let me recount.

Wait — total 11 numbers. After removing median (position 6), we have:

Positions 1-5: lower half → 6, 7, 15, 36, 39 → Q1 = median of these = 3rd number = 15

Positions 7-11: upper half → 41, 43, 43, 47, 49 → Q3 = median of these = 3rd number = 43

Wait — position 7 is 41? Let me write positions clearly:

Ordered data with positions:

1: 6
2: 7
3: 15
4: 36
5: 39
6: 41 ← median
7: 41
8: 43
9: 43
10: 47
11: 49

So lower half: positions 1-5 → [6,7,15,36,39] → median = 15 → Q1 = 15

Upper half: positions 7-11 → [41,43,43,47,49] → median = 43 → Q3 = 43

✔ So:

Ordered data: 6, 7, 15, 36, 39, 41, 41, 43, 43, 47, 49
Median: 41
Q3: 43
Q1: 15

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Problem 3: Angela’s sales records

Data: 34, 47, 1, 15, 57, 24, 20, 11, 19, 50, 28, 37

First, order the data:

1, 11, 15, 19, 20, 24, 28, 34, 37, 47, 50, 57

There are 12 numbers (even count).

a) Median = average of 6th and 7th numbers
6th = 24, 7th = 28 → (24+28)/2 = 52/2 = 26

b) Range = largest - smallest = 57 - 1 = 56

c) Upper and lower quartiles

Split into two halves (since even, no need to exclude middle):

Lower half: first 6 numbers → 1, 11, 15, 19, 20, 24
→ Q1 = median of lower half = average of 3rd and 4th = (15+19)/2 = 34/2 = 17

Upper half: last 6 numbers → 28, 34, 37, 47, 50, 57
→ Q3 = median of upper half = average of 3rd and 4th = (37+47)/2 = 84/2 = 42

d) Interquartile range (IQR) = Q3 - Q1 = 42 - 17 = 25

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Problem 4: Heights of 14 students

Heights: 65, 63, 68, 59, 74, 59, 68, 61, 64, 60, 69, 72, 55, 64

a) Complete the frequency table.

Intervals:

55-58: look for heights ≥55 and ≤58 → 55 → count = 1

59-62: 59, 59, 60, 61 → count = 4

63-66: 63, 64, 64, 65 → count = 4

67-70: 68, 68, 69 → count = 3

71-74: 72, 74 → count = 2

Check total: 1+4+4+3+2 = 14 ✔

Table:

| Interval | Frequency |
|----------|-----------|
| 55-58 | 1 |
| 59-62 | 4 |
| 63-66 | 4 |
| 67-70 | 3 |
| 71-74 | 2 |

b) Which interval contains the median?

Total 14 students → median is average of 7th and 8th values when ordered.

Order all heights:

55, 59, 59, 60, 61, 63, 64, 64, 65, 68, 68, 69, 72, 74

7th = 64, 8th = 64 → median = 64

Which interval contains 64? → 63-66 → 63-66

c) Which interval contains the upper quartile?

Q3 is the median of the upper half.

Since n=14, upper half is last 7 numbers: positions 8 to 14 → 64, 65, 68, 68, 69, 72, 74

Median of these = 4th number → 68

68 is in interval 67-70

d) What percent of students are shorter than 5 feet 7 inches?

5 feet 7 inches = 5×12 + 7 = 60 + 7 = 67 inches

Shorter than 67 → less than 67

Count how many heights < 67:

From ordered list: 55, 59, 59, 60, 61, 63, 64, 64, 65 → that’s 9 students

(68 and above are not included)

Total students = 14

Percent = (9 / 14) × 100 ≈ ?

Calculate: 9 ÷ 14 = 0.642857... × 100 ≈ 64.29%

We can round to nearest whole number or keep one decimal? Since it's money or measurement, maybe one decimal.

But let’s see what’s expected. Probably acceptable as fraction or rounded.

Actually, 9/14 * 100 = 900/14 = 450/7 ≈ 64.2857... → so 64.3% if rounding to one decimal.

But sometimes they want exact fraction. However, question says “percent”, so decimal is fine.

Alternatively, leave as 64.29% but I think 64.3% is standard.

Wait — let me double-check count:

Heights less than 67:

55, 59, 59, 60, 61, 63, 64, 64, 65 → yes, 9

Yes.

So percent = (9/14)*100 = 64.2857...% → typically rounded to 64.3%

But some teachers prefer exact fraction. However, since it says "percent", decimal is appropriate.

I’ll go with 64.3%

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Final Answers:

Problem 1:
a) 32
b) 27
c) 3.9

Problem 2:
Ordered data: 6, 7, 15, 36, 39, 41, 41, 43, 43, 47, 49
Median: 41
Q3: 43
Q1: 15

Problem 3:
a) Median: 26
b) Range: 56
c) Q1: 17, Q3: 42
d) IQR: 25

Problem 4:
a) Frequencies: 1, 4, 4, 3, 2
b) 63-66
c) 67-70
d) 64.3%

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Final Answer:
1. a) 32, b) 27, c) 3.9
2. Ordered data: 6, 7, 15, 36, 39, 41, 41, 43, 43, 47, 49; Median: 41; Q3: 43; Q1: 15
3. a) 26, b) 56, c) Q1=17, Q3=42, d) 25
4. a) Frequencies: 1, 4, 4, 3, 2; b) 63-66; c) 67-70; d) 64.3%