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