Answer:
Kindly check explanation
Step-by-step explanation:
Given the data:
38, 50, 50, 55, 55, 95, 95, 130, 180, 213, 250, 350, 450, 1750, 3000
The range :
(maximum - minimum) = (3000 - 38) = 2962
Variance :
Σ(x - mean)² / (n-1)
Where n = number of observations
Mean = (38+50+50+55+55+95+95+130+180+213+250+350+450+1750+3000) = 6761 / 15 = $450.73
Σ(x - mean)² = 9527804.9333334
N - 1 = 15 - 1 = 14
Hence,
Variance = Σ(x - mean)² / (n-1)
Variance = 9527804.9333334 / 14
Variance = $680557.50
Standard deviation = √variance
Standard deviation = √680557.50
Standard deviation = $824.96
CHECK IF OUTLIERS EXIST:
Using the outlier formula:
Lower outlier = Q1 - (1.5 × IQR)
Upper outlier = Q3 + (1.5 × IQR)
Q1 = 55 ; Q3 = 350 ; IQR = 295
Lower outlier = 55 - (1.5 × 295) = - 387.5
Upper outlier = 350 + (1.5 × 295) = 792.5
Remove values below the lower outlier value and values above the upper outlier value.
Hence outliers in the data are : 1750 and 3000
To save computation time :
We can recalculate our statistical measure by leaving out the outliers using online, excel or calculator functions :
Data without outlier :
38, 50, 50, 55, 55, 95, 95, 130, 180, 213, 250, 350, 450
Range = (450 - 38) = 412
Variance = Σ(x - mean)² / (n-1)
Variance = 201626.77 / 12
Variance = $16,802.23
Standard deviation:
√variance
√16,802.23
= $129.62341
According to the results obtained the presence of outliers has a great effect on the outcome the result.
