Question

2.
42, 38, 45, 68, 40, 39, 39, 41
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 check the data set: \(42, 38, 45, 68, 40, 39, 39, 41\). The value \(68\) is much larger than the others, so it's the outlier.

Step2: Calculate with outlier (all 8 data points)

Mean:

Sum of data: \(42 + 38 + 45 + 68 + 40 + 39 + 39 + 41 = 352\)
Mean = \(\frac{352}{8} = 44\)

Median:

Sort the data: \(38, 39, 39, 40, 41, 42, 45, 68\)
There are 8 values, so median is the average of the 4th and 5th: \(\frac{40 + 41}{2} = 40.5\)

Mode:

The value \(39\) appears twice, others once. So mode is \(39\)

Range:

Max - Min = \(68 - 38 = 30\)

Stdevp (Population Standard Deviation):

First, find the mean (\(\mu = 44\))
Deviations squared: \((42 - 44)^2 = 4\), \((38 - 44)^2 = 36\), \((45 - 44)^2 = 1\), \((68 - 44)^2 = 576\), \((40 - 44)^2 = 16\), \((39 - 44)^2 = 25\), \((39 - 44)^2 = 25\), \((41 - 44)^2 = 9\)
Sum of squared deviations: \(4 + 36 + 1 + 576 + 16 + 25 + 25 + 9 = 692\)
Stdevp = \(\sqrt{\frac{692}{8}} = \sqrt{86.5} \approx 9.3\) (rounded to one decimal)

Step3: Calculate without outlier (remove 68, 7 data points: \(42, 38, 45, 40, 39, 39, 41\))

Mean:

Sum of data: \(42 + 38 + 45 + 40 + 39 + 39 + 41 = 284\)
Mean = \(\frac{284}{7} \approx 40.57\) (rounded to two decimals)

Median:

Sort the data: \(38, 39, 39, 40, 41, 42, 45\)
There are 7 values, so median is the 4th value: \(40\)

Mode:

The value \(39\) still appears twice, others once. So mode is \(39\)

Range:

Max - Min = \(45 - 38 = 7\)

Stdevp (Population Standard Deviation):

Mean (\(\mu = \frac{284}{7} \approx 40.57\))
Deviations squared: \((42 - 40.57)^2 \approx 2.04\), \((38 - 40.57)^2 \approx 6.60\), \((45 - 40.57)^2 \approx 19.62\), \((40 - 40.57)^2 \approx 0.32\), \((39 - 40.57)^2 \approx 2.46\), \((39 - 40.57)^2 \approx 2.46\), \((41 - 40.57)^2 \approx 0.18\)
Sum of squared deviations: \(\approx 2.04 + 6.60 + 19.62 + 0.32 + 2.46 + 2.46 + 0.18 = 33.68\)
Stdevp = \(\sqrt{\frac{33.68}{7}} \approx \sqrt{4.81} \approx 2.19\) (rounded to two decimals)

Step4: Calculate Lower and Upper Bounds (for outlier detection, using IQR method)

First, for data without outlier (sorted: \(38, 39, 39, 40, 41, 42, 45\))
Q1 (25th percentile): median of lower half (\(38, 39, 39\)) = \(39\)
Q3 (75th percentile): median of upper half (\(41, 42, 45\)) = \(42\)
IQR = \(Q3 - Q1 = 42 - 39 = 3\)
Lower Bound = \(Q1 - 1.5 \times IQR = 39 - 4.5 = 34.5\)
Upper Bound = \(Q3 + 1.5 \times IQR = 42 + 4.5 = 46.5\) (this shows 68 is above upper bound, confirming it's an outlier)

Answer:

With Outlier:

• Mean: \(44\)
• Median: \(40.5\)
• Mode: \(39\)
• Range: \(30\)
• Stdevp: \(\approx 9.3\)

Without Outlier:

• Mean: \(\approx 40.57\)
• Median: \(40\)
• Mode: \(39\)
• Range: \(7\)
• Stdevp: \(\approx 2.19\)
• Lower Bound: \(34.5\)
• Upper Bound: \(46.5\)