In: Statistics and Probability
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 64
Arizona 515
Maryland, Delaware, DC 303
Connecticut 191
Florida 715
Georgia 232
Hawaii 95
Maine 76
Minnesota, Dakotas 286
Missouri, Kansas 349
Mississippi 55
Nebraska 78
North Carolina 734
Nevada 113
New Jersey, Bermuda 328
Alaska, Idaho, Montana,Oregon, Washington
875
New York 1,075
Ohio 623
Oklahoma 112
Arkansas 51
Illinois 595
Indiana 315
Iowa 93
Kentucky 130
Louisiana 301
Michigan 413
Massachusetts 343
California 939
New Mexico 105
Pennsylvania 392
Rhode Island 30
Colorado 534
South Carolina 175
Texas 999
Tennessee 238
Utah 50
Virginia 498
Vermont, New Hampshire 104
Wisconsin 430
West Virginia 62
Click here for the Excel Data File
Find the mean, median, and standard deviation of the number of
members per component. (Round your answers to 2 decimal
places.)
b-1. Find the coefficient of skewness, using the software method. (Round your answer to 2 decimal places.)
b-2. What do you conclude about the shape of the distribution of component size?
Mild positive skewness
Mild negative skewness
Determine the first and third quartiles. Do not use the method
described by Excel. (Round your answers to 2 decimal
places.)
d-1. Are there any outliers?
Zero
One
Three
Four
Two
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.)
Please refer below provided Minitab output:
To get the output
Open Minitab - Go to Stat - Basic Satistics - Display Descriptive Statistics - In Satistics option choose the terms you want to calculate.
a) Mean = 340.4, Median = 293.5, Standard Deviation = 289.3
b) Coefficient of Skewness = 1.05
c) Since, Mean > Median, the shape of distribution is positively skewed.
d) The First and Third quartiles will be:
Q1 = 97.3 and Q3 = 510.8
e) There are five outliers.
f) The limits of outliers can be calculated as below:
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