Answer:
a) 3.128
b) Yes, it is an outerlier
Step-by-step explanation:
The standardized z-score for a particular sample can be determined via the following expression:
z_i = {x_i -\bar x}/{s}
Where;
\bar x = sample means
s = sample standard deviation
Given data:
the mean shipment thickness (\bar x) = 0.2731 mm
With the standardized deviation (s) = 0.000959 mm
The standardized z-score for a certain shipment with a diameter x_i= 0.2761 mm can be determined via the following previous expression
z_i = {x_i -\bar x}/{s}
z_i = {0.2761-0.2731}/{ 0.000959}
z_i = 3.128
b)
From the standardized z-score
If [z_i < 2]; it typically implies that the data is unusual
If [z_i > 2]; it means that the data value is an outerlier
However, since our z_i > 3 (I.e it is 3.128), we conclude that it is an outerlier.
Answer:
Option C
Step-by-step explanation:
Just took the quiz and it was right.
Answer:
Divide StartFraction a Over 11 EndFraction by 11; the solution is 540.
Step-by-step explanation:
Answer:
In order, they would be:
-4.45 min, -1.8 min, -(1 + 2/3) min, -1.375 min
Step-by-step explanation:
Enjoy :)
Answer:
Step-by-step explanation:
No there parallel
