Your answer would be -2.2
No. If she selects a breakfast at random there is an equal chance she will select each item. 25% chance she will select oatmeal, 25% chance she will select cereal, 25% she will select french toast, and a 25% chance she will select scrambled eggs.
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.
we know that the multiplication comes first in the family then we start from left to right
4 x 5= 20
9-20 = -11
-11 + 6 = -5
there you go!
