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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: S-ID.A.1, S-ID.A.2, S-ID.A.3 · Level: Grade 9 Algebra 1
Problem 2 Median, Odd Count
Sorted: 36, 38, 39, 41, 43, 45, 50. The middle (4th) value is 41.
41
Problem 3 Median, Even Count
Sorted: 7, 9, 10, 11, 12, 15. The two middle values are 10 and 11, and their mean is 10.5.
10.5
Problem 4 Range and IQR from a Summary
Range: 52 − 14 = 38. IQR: 35 − 22 = 13.
Range 38; IQR 13
Problem 5 Read a Dot Plot
The data are 6, 7, 7, 7, 8, 8, 8, 8, 9, 9, 10, 12. The tallest stack is at 8, so the mode is 8. The 6th and 7th values are both 8, so the median is 8. Range: 12 − 6 = 6.
Mode 8; median 8; range 6
Problem 6 Read a Histogram
The tallest bar is 20 to 40 (8 students). The last two bars are 60 to 80 (2) and 80 to 100 (1): 3 students. 3 ÷ 20 = 0.15.
20 to 40 minutes; 3 students; 15%
Problem 7 Read a Box Plot
Range: 40 − 5 = 35. IQR: 24 − 12 = 12. The box holds the middle half of the data: 50%.
Outlier fence: 24 + 1.5(12) = 42. Since 40 < 42, it is not an outlier.
Range 35; IQR 12; 50%; 40 is not an outlier
Problem 8 Five-Number Summary, Odd Count
The data are sorted and n = 11. The median is the 6th value, 12. Lower half: 4, 6, 7, 9, 10 gives Q1 = 7. Upper half: 13, 15, 18, 20, 26 gives Q3 = 18.
IQR = 18 − 7 = 11.
Min 4, Q1 7, median 12, Q3 18, max 26; IQR 11
Problem 9 Five-Number Summary and Outliers
n = 10. Median: (30 + 31) ÷ 2 = 30.5. Lower half 21, 24, 24, 27, 30 gives Q1 = 24. Upper half 31, 33, 35, 39, 58 gives Q3 = 35. IQR = 11.
Upper fence: 35 + 1.5(11) = 51.5. Lower fence: 24 − 16.5 = 7.5. Only 58 lies outside.
Min 21, Q1 24, median 30.5, Q3 35, max 58; the value 58 is an outlier
Problem 10 Outlier and the Mean
Before: mean 86 ÷ 5 = 17.2; median 17. After: 14 + 16 + 17 + 19 + 70 = 136, so the mean is 27.2; the median is still 17.
The mean rose by 10, but the median did not change.
Mean 17.2 → 27.2; median 17 → 17
Problem 11 Compare Spread with Standard Deviation
Both means are 20 ÷ 4 = 5. Set A: squared distances 4, 0, 0, 4 sum to 8; variance 8 ÷ 4 = 2; σ = √2 ≈ 1.41.
Set B: 16, 0, 0, 16 sum to 32; variance 8; σ = √8 ≈ 2.83.
Mean 5 for both; σ ≈ 1.41 for A and ≈ 2.83 for B, so B is more spread out
Problem 12 Mean or Median?
Mean: (210 + 225 + 230 + 240 + 250 + 1900) ÷ 6 = 3055 ÷ 6 ≈ 509.2. Median: (230 + 240) ÷ 2 = 235.
The 1,900 pulls the mean far above every other price.
Mean about $509,200; median $235,000; the median is the better description
Problem 13 Describe a Histogram
Most students sent 5 to 15 messages, and a long tail extends to the right: the distribution is skewed right.
The few high values pull the mean up, so the mean is greater than the median.
Skewed right; mean greater than median
Problem 14 Compare Two Box Plots
Medians: A 22, B 24. IQRs: A 27 − 18 = 9; B 26 − 20 = 6. Ranges: A 35 − 12 = 23; B 40 − 8 = 32.
Team B has the higher median and the smaller IQR, so its typical games are more consistent, even though one extreme game makes its range larger.
B: higher median, smaller IQR (6 vs 9); more consistent in the middle half
Problem 15 Mean from a Frequency Table
Total pets: 0(5) + 1(8) + 2(4) + 3(2) + 4(1) = 0 + 8 + 8 + 6 + 4 = 26. Mean: 26 ÷ 20 = 1.3.
The 10th and 11th values lie in the group of 1 pet (cumulative counts 5, then 13). The median is 1.
Mean 1.3 pets; median 1 pet
Problem 16 Find the Missing Score
Five quizzes need a total of 5 × 86 = 430. The first four add to 341.
430 − 341 = 89.
89
Problem 17 Adding a Seventh Value
The original sum is 6 × 14 = 84. New sum: 84 + 28 = 112. New mean: 112 ÷ 7 = 16.
A new value equal to the current mean, 14, adds 14 to the sum and 1 to the count, so the mean stays 14.
New mean 16; adding 14 leaves it unchanged
Problem 18 Two Bakeries
Medians: X 60, Y 58, so X is higher. IQRs: X 70 − 52 = 18; Y 62 − 55 = 7, so Y is more consistent.
Fence for X: 70 + 1.5(18) = 97. Since 95 < 97, there is no outlier.
X has the higher median; Y is more consistent; no outlier in X
Problem 19 Consistent Runner
Both means are 25. P: squared distances 1, 0, 1 sum to 2; variance 2 ÷ 3; σ ≈ 0.82. Q: 9, 0, 9 sum to 18; variance 6; σ ≈ 2.45.
P has the smaller standard deviation, so the times of P cluster more tightly around 25 minutes.
σ ≈ 0.82 for P and ≈ 2.45 for Q; P is more consistent
Problem 20 Shifting and Scaling Data
Original mean: 5.
(a) The set becomes 12, 14, 16, 18. The mean is 15, and the distances from the mean do not change, so σ = √5 ≈ 2.24.
(b) The set becomes 4, 8, 12, 16. The mean is 10, and every distance doubles, so σ = 2√5 ≈ 4.47.
Adding a constant shifts the center only; multiplying scales both the center and the spread.
(a) mean 15, σ ≈ 2.24; (b) mean 10, σ ≈ 4.47