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:
c is the answer
Step-by-step explanation:
Answer:
NO
Step-by-step explanation:
bc
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home
work
First of all, you should know that:
direct variation:
y = c * x
Inverse variation:
y = c / x
where, c is any number that belongs to the real ones.
answer:
An equation that represents an inverse variation is
y = 4 / x
Answer:
-10
Step-by-step explanation:
(1/5)x+4=2
5(1/5)x+4=2)
x+20=10
x = -10
