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.
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Answer: try the second one
Step-by-step explanation:
Y-4= -3(x-3)
First you'll distribute the -3(x-3) and you should get -3x+9 (a negative divided by a negative is a positive)
So your new equation will be y-4=-3x+9. Next you'll get rid of the -4 y adding 4 on both sides of the equation. So you should get y=-3x+13. So points you could use is <span><span>(<span>0,13</span>)</span>,<span>(<span>1,10</span>)</span>,<span>(<span>2,7</span>)</span></span><span />
