Listed below are the amounts (dollars) it costs for marriage proposal packages at different baseball stadiums. Find the range, variance, and standard deviation for the given sample data. Include appropriate units in the results. Are there any outliers, and are they likely to have much of an effect on the measures of variation? 38, 50, 50, 55, 55, 95, 95, 130, 180, 213, 250, 350, 450, 1750, 3000

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.

the least common denominator is 40

by definition, an arc subtended by a central angle of some given measure has the same measure. so .