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.
For the first one it is 5 x10^{8}
and for the second one it is 0.0005 = 5 × 0.0001 = 5 × 10−4
The answer that you are looking for is 2
Answer:
27 oak trees
Step-by-step explanation:
X is equal 25 degrees. Line ac goes though the parallel lines. So bcd is x. Cdb is 180-50 equals 130. All angles add to 180. 180-130 equals 2X. So x equals 25.
