shaded arrow to the left.
shaded in circle.
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.
A jar of jelly beans contains 50 red gumballs , 45 yellow gumballs, and 30 green gumballs. You reach into the jar and randomly
select a jelly bean, then select another without
putting the first jelly bean back. What is the
probability that you draw two red jelly beans? This is Dependent because you didnt put the other jelly bean in thus changing the total nmber of jelly beans.
A jar of jelly beans contains 50 red gumballs<span> , 45 yellow gumballs, and 30 green gumballs. You reach into the jar and randomly select a jelly bean, then select another while replacing the first jelly bean back. What is the probability that you draw two red jelly beans? This is Independent because you put the other jelly bean in thus keeping the total number of jelly beans.</span>
