Answer:
The probability of finding aleast one or more defective charmins = 0.40951
Step-by-step explanation:
The probability of not finding a defective charmin = 1
probability of finding defective charmain = 1-0.1 =0.9
The probability of not finding all five defective item =
Then,
probability of not finding all five defective item=0.59049
Thus,
The probability of finding atleast one or more defective chairmins = 1-0.59049
Therefore,
The probability of finding atleast one or more defective chairmins =0.40951
Answer:
25%
Step-by-step explanation:
|×100
×100
=0.25×100
=25%
=25%error
That outlier will appear as a single blip of data, but it will be far
away from the central part of the histogram. It will not be that visible
if there is a lot of data, since 1 outlying data point will just show a
very low height compared to the central data near the mean/median/mode,
where there will be higher frequencies and higher bars of data.
If these are the choices:
<span>A.The outlier will appear as a tall bar near one side of the distribution.
B.The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.
C.Since a histogram shows frequencies, not individual data values, the
outlier will not appear. Instead, the outlier increases the frequency
for its class by 1.
D.The outlier will appear as the tallest bar near the center of the distribution.
</span>
Then the most accurate answer will be B, since it's just a single value with a bar corresponding to a frequency of 1.
