What is the outlier in the following data set:<br>
15,11,10,8,9,1,8,7,5,4,2,3, and 37?
NeTakaya
Step-by-step explanation:
The steps to find an outlier:
1. Put the data in numerical order.
2. Find the median.
3. Find the medians for the top and bottom parts of the data. This divides the data into 4 equal parts.
The median with the smallest value is called Q1. The median for all the values - usually just called the median is also called Q2. The median with the largest value is Q3.
4. Subtract...Q3 - Q1. This value is the InterQuartileRange or IQR. Remember that the range means taking the largest minus the smallest. This is a special range having to do with the quartiles.
5. Multiply...1.5 * IQR
6. Take your answer from #5 and do 2 things with it. A). Subtract it from Q1 and B) Additional to Q3.
7. Look at all your data points. If any are SMALLER than Q1 - 1.5 *IQR, they are outliers. If any are LARGER than Q3 + 1.5 *IQR, they are also outliers.
For your data....the median, Q2 is
(43+38)/2 = 40.5.
Q1 = (30+26)/2 = 28.
Q3 = (54+52)/2 = 53
The IQR is 53 - 28 = 25
1.5 * IQR = 37.5
Q1 - 37.5 = 28 - 37.5 = -9.5. There is no data value less than -9.5.
Q3 + 37.5 = 53 + 37.5 = 90.5. there is no data value greater than 90.5.
My conclusion is that there are no outliers in this data.
I hope this helps!
Your first step is to do 6(x) which is 6 multiplied by x which is the value of 5\6
Answer:
a) C. No, a carton can have a puncture and a smashed corner.
b) The probability that a carton has a puncture or a smashed corner is P(X ∪ Y) = 0.104.
Step-by-step explanation:
To be mutually exclusive, the probability of the two events happening at the same time should be 0. But the probability that a carton has a puncture and has a smashed corner is 0.004 and not 0.
Then, we can conclude the events "selecting a carton with a puncture" and "selecting a carton with a smashed corner" are not mutually exclusive.
The answer is "C. No, a carton can have a puncture and a smashed corner."
We can calculate the probability that the carton has a puncture <em>or </em>has a smashed comer simply by adding the probability of each event:
P(X ∪ Y): probability that a carton has a puncture or a smashed corner.
2x+4y=16
x-7=-4y
4y=16-2x
<span>4y=16-2(-4y+7)
4y=16+8y-14
-4y=2
-4y/-4=2/-4
<u>y=-0,5</u>
2x=16-4(-0,5)
2x=16+2
2x/2=18/2
<u>x=9
</u>
C</span>
Assuming the n data values are independent draws from a Poisson distributed population with parameter λ , the MLE of λ from the sample is the sample mean, i.e.
Numerous proves are available online.
