Question

The American Society of PeriAnesthesia Nurses (ASPAN; www.aspan.org) is a national organization serving nurses practicing in ambulatory surgery, preanesthesia, and postanesthesia care. The organization's membership is listed below.

State/Region	Membership
Alabama		102
Arizona		420
Maryland, Delaware, DC		471
Connecticut		177
Florida		432
Georgia		330
Hawaii		84
Maine		51
Minnesota, Dakotas		360
Missouri, Kansas		325
Mississippi		128
Nebraska		77
North Carolina		368
Nevada		121
New Jersey, Bermuda		508
Alaska, Idaho, Montana,Oregon, Washington		716
New York		1,016
Ohio		833
Oklahoma		209
Arkansas		89
Illinois		492
Indiana		332
Iowa		83
Kentucky		236
Louisiana		290
Michigan		413
Massachusetts		540
California		1,173
New Mexico		62
Pennsylvania		443
Rhode Island		43
Colorado		443
South Carolina		281
Texas		1,168
Tennessee		169
Utah		88
Virginia		412
Vermont, New Hampshire		109
Wisconsin		471
West Virginia		77
a. Find the mean, median, and standard deviation of the number of members per component. (Round your answers to 2 decimal places.) Mean: Median: Standard Deviation: b-1. Find the coefficient of skewness, using the software method. (Round your answer to 2 decimal places.) Coefficient of skewness: b-2. What do you conclude about the shape of the distribution of component size? a. Mild positive skewness b. Mild negative skewness c. Determine the first and third quartiles. Do not use the method described by Excel. (Round your answers to 2 decimal places.) First Quartile: Third Quartile: d-1. Are there any outliers? a. Three b. One c. Two d. Four e. Zero d-2. What are the limits for outliers? (Round your answers to the nearest whole number. Negative amounts should be indicated by a minus sign.) Limits _____ to _____

Answer 1

a)

Mean = Total of Values/Count of Values = 14142/40 = 353.55

Median is the middle value when arranged in ascending order. Since the total count is even, Median will be the average of the two middle values.

Median = 327.5

Standard deviation:

Using the above formulae,

Standard Deviation of the data(σ) = 292.01

B1)

Skewness = 1.39

B2)

A distribution is skewed if one of its tails is longer than the other. Since our histogram has a long tail in the positive direction it is mild positively skewed.

c)

Steps to find Quartile 1 and 3

• First, we write data in increasing order
• Find the median
• the lower half of the data is minimum value till median
• The first quartile, Q₁, is the median of the values of lower half
• Similarly, upper half of the data is median till maximum value
• The third quartile, Q₃, is the median of the values of upper half

Q1 = 107.25

Q3 = 450

d2)

Inter Quartile Range (IQR) = Q3-Q1 = 342.75

Range to identify Outliers ie {Q1-1.5*IQR, Q3+1.5*IQR} ie

{-406.88,964.13}

Limits: -406.88 to 964.13

d1)

Since 3 data points are outside this interval, hence there are 3 outliers.