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Question
section ii: free response
- below are data for 20 students quiz scores:
4, 5, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 10, 10, 10, 11, 12, 13, 14, 18
(a) construct a histogram with bin width 3.
(b) are there any outliers? justify.
(c) should mean/sd or median/iqr be used? explain.
- a population is 120,000 people and grows at 3% per year.
(a) write an exponential growth equation.
(b) predict population in 10 years.
Problem 1(a)
Step1: Determine bins
The minimum score is 4, maximum is 18. Bin width = 3. So bins are: \(4 - 6\), \(7 - 9\), \(10 - 12\), \(13 - 15\), \(16 - 18\).
Step2: Count frequencies
- \(4 - 6\): Scores 4,5,6,6 → Frequency = 4
- \(7 - 9\): Scores 7,7,7,8,8,8,9,9 → Frequency = 8
- \(10 - 12\): Scores 10,10,10,11,12 → Frequency = 5
- \(13 - 15\): Scores 13,14 → Frequency = 2
- \(16 - 18\): Score 18 → Frequency = 1
Step3: Draw histogram
- X - axis: Bins (\(4 - 6\), \(7 - 9\), \(10 - 12\), \(13 - 15\), \(16 - 18\))
- Y - axis: Frequency (0 - 8, scale as needed)
- Bars: Heights = frequencies (4,8,5,2,1) for respective bins.
Step1: Find Q1, Q3, IQR
Data sorted: 4,5,6,6,7,7,7,8,8,8,9,9,10,10,10,11,12,13,14,18. \(n = 20\).
- Q2 (median): Average of 10th and 11th terms: \(\frac{8 + 9}{2}=8.5\)
- Q1: Median of first 10 terms (4,5,6,6,7,7,7,8,8,8) → \(\frac{7 + 7}{2}=7\)
- Q3: Median of last 10 terms (9,9,10,10,10,11,12,13,14,18) → \(\frac{10 + 11}{2}=10.5\)
- IQR = Q3 - Q1 = \(10.5 - 7 = 3.5\)
Step2: Check outliers
Lower fence: \(Q1 - 1.5\times IQR = 7 - 5.25 = 1.75\)
Upper fence: \(Q3 + 1.5\times IQR = 10.5 + 5.25 = 15.75\)
Score 18 > 15.75 → Outlier.
Step1: Analyze data shape
There is an outlier (18) and data may be skewed (right - skewed due to 18).
Step2: Choose measures
Mean/SD are sensitive to outliers. Median/IQR are resistant to outliers. So use median/IQR.
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Histogram with bins \(4 - 6\) (freq=4), \(7 - 9\) (freq=8), \(10 - 12\) (freq=5), \(13 - 15\) (freq=2), \(16 - 18\) (freq=1)