Question

1. 76, 35, 37, 38, 3, 40, 44, 42

with outlier
mean:
median:
mode:
range:
stdevp:
without outlier
mean;
median:
mode:
range:
stdevp:
lower bound:
upper bound:

Explanation:

Step1: Identify the outlier

First, we need to find the outlier. Let's sort the data: \(3, 35, 37, 38, 40, 42, 44, 76\). The outlier is likely \(3\) or \(76\) as they are far from the other values. Let's use the interquartile range (IQR) method. First, find the median of the data to split into lower and upper halves. The data has 8 values, so the median is the average of the 4th and 5th values: \(\frac{38 + 40}{2}=39\). The lower half is \(3, 35, 37, 38\), median (Q1) is \(\frac{35 + 37}{2}=36\). The upper half is \(40, 42, 44, 76\), median (Q3) is \(\frac{42 + 44}{2}=43\). IQR = Q3 - Q1 = \(43 - 36 = 7\). Lower bound for outliers: \(Q1 - 1.5\times IQR = 36 - 10.5 = 25.5\). Upper bound: \(Q3 + 1.5\times IQR = 43 + 10.5 = 53.5\). So \(76\) is above the upper bound and \(3\) is below the lower bound? Wait, \(3<25.5\), so \(3\) is an outlier? Wait, no, \(76>53.5\), so \(76\) is also an outlier? Wait, maybe the original data's outlier is \(3\) or \(76\). But looking at the "Without Outlier" column, the data should be without one outlier. Let's assume the outlier is \(3\) (since \(76\) is also far, but let's check the mean. With all data: sum is \(3 + 35 + 37 + 38 + 40 + 42 + 44 + 76 = 315\). Mean is \(315\div8 = 39.375\approx39.4\) (close to the written \(39.3\)). Without \(3\), sum is \(315 - 3 = 312\), number of data points is \(7\), mean is \(312\div7\approx44.57\) (close to the written \(44.6\)). Without \(76\), sum is \(315 - 76 = 239\), mean is \(239\div7\approx34.14\), which doesn't match the written \(44.6\). So the outlier is \(3\).

Step2: Calculate with outlier (all data)

- **Mean**: Sum all values: \(3 + 35 + 37 + 38 + 40 + 42 + 44 + 76 = 315\). Mean = \(\frac{315}{8}=39.375\approx39.4\) (close to the written \(39.3\)).
- **Median**: As calculated before, median is \(\frac{38 + 40}{2}=39\) (matches the written \(39\)).
- **Mode**: No value repeats, so mode is "no mode" (matches the written).
- **Range**: Max - Min = \(76 - 3 = 73\) (matches the written \(73\)).
- **Stdevp (population standard deviation)**: Formula is \(\sqrt{\frac{\sum (x_i - \mu)^2}{N}}\). \(\mu = 39.375\). Calculate each \((x_i - \mu)^2\):
- \((3 - 39.375)^2 = (-36.375)^2 = 1324.140625\)
- \((35 - 39.375)^2 = (-4.375)^2 = 19.140625\)
- \((37 - 39.375)^2 = (-2.375)^2 = 5.640625\)
- \((38 - 39.375)^2 = (-1.375)^2 = 1.890625\)
- \((40 - 39.375)^2 = (0.625)^2 = 0.390625\)
- \((42 - 39.375)^2 = (2.625)^2 = 6.890625\)
- \((44 - 39.375)^2 = (4.625)^2 = 21.390625\)
- \((76 - 39.375)^2 = (36.625)^2 = 1341.4140625\)
Sum of these squares: \(1324.140625 + 19.140625 + 5.640625 + 1.890625 + 0.390625 + 6.890625 + 21.390625 + 1341.4140625 = 2720.9375\). Then stdevp = \(\sqrt{\frac{2720.9375}{8}}=\sqrt{340.1171875}\approx18.44\) (matches the written \(18.44\)).

Step3: Calculate without outlier (remove 3)

Data becomes: \(35, 37, 38, 40, 42, 44, 76\)? Wait, no, if we remove \(3\), the data is \(35, 37, 38, 40, 42, 44, 76\)? Wait, but \(76\) is still an outlier? Wait, maybe the outlier is \(76\). Let's check. Sum with \(76\) is \(315\), remove \(76\), sum is \(315 - 76 = 239\), number of data points is \(7\). Mean is \(239\div7\approx34.14\), which doesn't match the written \(44.6\). Wait, the written "Without Outlier" mean is \(44.6\), so let's recalculate. Wait, maybe the outlier is \(3\), but then the data without \(3\) is \(35, 37, 38, 40, 42, 44, 76\) (7 points). Sum: \(35 + 37 + 38 + 40 + 42 + 44 + 76 = 312\)? Wait, no, \(35+37=72, +38=110, +40=150, +42=192, +44=236, +76=312\). \(312\div7\approx44.57\approx44.6\) (matches the written). So the outlier is \(3\). Now, data without \(3\): \(35, 37, 38, 40, 42, 44, 76\)? Wait, no, \(76\) is still there. Wait, maybe the outlier is \(3\) and \(76\) is not considered? Wait, maybe the original data's outlier is \(3\), and the "Without Outlier" data is \(35, 37, 38, 40, 42, 44\) (remove \(3\) and \(76\))? But that would be 6 data points. Sum: \(35+37+38+40+42+44=236\), mean \(236\div6\approx39.33\), not \(44.6\). Wait, there's a mistake in the written answers? Or maybe I misidentified the outlier. Wait, the written "Without Outlier" mean is \(44.6\), which is \(312\div7\approx44.57\approx44.6\), so the data without outlier is 7 points (remove \(3\)). Now, median: 7 data points, median is the 4th value. Sorted data: \(35, 37, 38, 40, 42, 44, 76\). Median is \(40\) (matches the written \(40\)). Mode: no mode (all values unique). Range: \(76 - 35 = 41\) (matches the written \(41\)). Stdevp: sum of squares for data \(35,37,38,40,42,44,76\). Wait, no, if we remove \(3\), the data is \(35,37,38,40,42,44,76\). Let's calculate stdevp. Mean is \(312\div7\approx44.57\). Squared deviations: \((35 - 44.57)^2\approx(-9.57)^2\approx91.58, (37 - 44.57)^2\approx(-7.57)^2\approx57.30, (38 - 44.57)^2\approx(-6.57)^2\approx43.16, (40 - 44.57)^2\approx(-4.57)^2\approx20.88, (42 - 44.57)^2\approx(-2.57)^2\approx6.60, (44 - 44.57)^2\approx(-0.57)^2\approx0.32, (76 - 44.57)^2\approx(31.43)^2\approx987.84\). Sum of squares: \(91.58 + 57.30 + 43.16 + 20.88 + 6.60 + 0.32 + 987.84 = 1207.68\). Stdevp = \(\sqrt{\frac{1207.68}{7}}\approx\sqrt{172.5257}\approx13.14\) (close to the written \(13.14\)).

Step4: Lower and Upper Bound (for outlier detection)

We calculated earlier with Q1=36, Q3=43, IQR=7. Lower bound: \(Q1 - 1.5\times IQR = 36 - 10.5 = 25.5\). Upper bound: \(Q3 + 1.5\times IQR = 43 + 10.5 = 53.5\).

Answer:

• With Outlier:
• Mean: \(39.375\) (or \(39.4\))
• Median: \(39\)
• Mode: No Mode
• Range: \(73\)
• Stdevp: \(18.44\)
• Without Outlier (remove 3):
• Mean: \(44.57\) (or \(44.6\))
• Median: \(40\)
• Mode: No Mode
• Range: \(41\)
• Stdevp: \(13.14\)
• Lower Bound: \(25.5\)
• Upper Bound: \(53.5\)