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.
1/9
2
1 4/5
3 1/2
There's your answer.
Answer:
it is 3,540ml that is less than 10 L
Step-by-step explanation:
Answer:
The standard, or near-average, atmospheric pressure at sea level on the Earth is 1013.25 millibars, or about 14.7 pounds per square inch.
The gauge pressure in my automobile tires is a little more than twice that value. If you live at a higher altitude, the pressure will be lower since there is less air above weighing down upon you. The pressure also varies by relatively small amounts as high and low pressure systems move past.
Step-by-step explanation:
