Answer:
a) 
b) 
c) 
With a frequency of 4
d) 
<u>e)</u>
And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:

Replacing we got:

Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

Part c
The mode is the most repeated value in the sample and for this case is:

With a frequency of 4
Part d
The midrange for this case is defined as:

Part e
For this case we can calculate the deviation given by:

And replacing we got:

And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
I am doing with Question Number 2
Let the two numbers be 'x' and 'y'.
According to the question, the product of two numbers is 14.

So, xy=14 (Equation 1)
Now, one of the numbers is 3 ½ times the other.
So, 
(Equation 2)
Substituting the value of 'x' from equation 2 in equation 1, we get



So, y= 2 and -2
So,
and 
x=7 and -7
Numbers are 2,7 and -2,-7.
Sum of numbers 2 and 7 = 9
Sum of numbers -2 and -7 = -9.