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 114
Arizona 261
Maryland, Delaware, DC 319
Connecticut 183
Florida 654
Georgia 313
Hawaii 58
Maine 98
Minnesota, Dakotas 335
Missouri, Kansas 324
Mississippi 52
Nebraska 125
North Carolina 342
Nevada 66
New Jersey, Bermuda 321
Alaska, Idaho, Montana, Oregon, Washington 893
New York 690
Ohio 798
Oklahoma 201
Arkansas 74
Illinois 632
Indiana 216
Iowa 76
Kentucky 155
Louisiana 161
Michigan 304
Massachusetts 530
California 1,110
New Mexico 93
Pennsylvania 631
Rhode Island 75
Colorado 344
South Carolina 264
Texas 1,120
Tennessee 122
Utah 45
Virginia 304
Vermont, New Hampshire 192
Wisconsin 316 West Virginia 63
Use statistical software to answer the following questions.
a. Find the mean, median, and standard deviation of the number of members per component. (Round your answers to 1 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?
Mild positive skewness or Mild negative skewness
(d-1) Are there any outliers?
One, Zero, Three, Four, Two
(d-2) What are the limits for outliers? (Round your answers to 1 decimal place. Negative amounts should be indicated by a minus sign.)

Answer 1

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data :

114, 261, 319, 183,654,313,58,98,335,324,52,125,342,66,321,893,690,798,201,74,632,216,76,155,161,304,530,1110,93,631,75,344,264,1120,122,45,304,192,316,63

USING CALCULATOR :

The mean(m) of the data: = ΣX/n

n = sample size = 40

m = 12974/40

m = 324.4

Median = ((n + 1)/2) th term = 41/2 = (20 + 21)th term / 2

= (261 + 264) / 2 = 262.5

Standard deviation (s) = sqrt(Σ(x - m)²/n))

s = 281. 74

B.) Coefficient of skewness (Ks) :

3(mean - median) / standard deviation

3(324.4 - 262.5) / 281.74

= 0.66

Mild positive skewness Given a positive skew Coefficient value.

(d-1) Are there any outliers?

Yes

Four outliers

(d-2) What are the limits for outliers? (Round your answers to 1 decimal place. Negative amounts should be indicated by a minus sign.)

Lower bound = Q1 - (1.5 * IQR)

Upper bound = Q3 + (1.5 * IQR)

IQR = Q3 - Q1

Q3 = upper quartile = 343 ; Q1 = 106 ; Q2 = 262.5

IQR = 343 - 106 = 237

Lower bound = 106 - (1.5 * 237) = - 249.5

Upper bound = 343 + (1.5 * 237) = 698.5

-249.5 ≤ X ≤ 698.5