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USPshnik [31]
3 years ago
6

Consider this example on how different a median can be from a sample. Suppose there is a company that mentions the average salar

y at Company A last year was $ 135,750 per year. Being interested in this you try to investigate this. You discover there are 8 individuals, including the business owner, that works at the company. You discover that the salaries of the 7 individuals are surprisingly low. The salaries of those 7 individuals are $8,000, $8,000, $7,000, $11,000, $15,000, $17,000, and $20,000. You then discover the salary of the business owner to be $1,000,000.
Compare the mean with the median? What does this show?
Find Q1 (first quartile) , Q3 (third quartile) , and Interquartile Range (IQR)?
Are there outliers? Please prove there is an outlier using the Q1-1.5*IQR and Q3 + 1.5*IQR formulas?
Find the 10% trimmed mean of the 8 salaries above. Compare this with the median found in part a.
What do you think represents a more typical salary of the organization above? Would you want to work for this company?
Mathematics
1 answer:
Delvig [45]3 years ago
5 0

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data :

8000, 8000, 7000, 11000, 15000, 17000, 20000, 1000000.

Mean = Σx / n = 1086000 / 8 = 135750

Median = 1/2(n+1)th term = 8000

The lower quartile :

Q1 = 1/4(n+1)th term

Q1 = 8000

Q3 = 3/4 (n+1)th term

Q3 = 18500

IQR = 18500 - 8000 = 10500

OUTLIER :

8000 - (1.5*(10500)) = - 7750

18500 + (1.5*(10500)) = 34250

10% trimmed mean

10% * 8

Cut off 1 from the top and bottom

Data becomes : 8000, 8000, 11000, 15000, 17000, 20000

Trimmed mean = 13166

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Solve for L: d= (LR_2)/(R_2+R_1
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D= (LR_2)/(R_2+R_1) 
<span>d(R_2+R_1)= LR_2 </span>
<span>d(R_2+R_1)/(R_2)= L </span>
<span>L= d(R_2+R_1)/(R_2) </span>

<span>The answer I obtained is the same answer as the third choice. </span>

<span>L= d(R_2+R_1)/(R_2) </span>
<span>L= (R_2+R_1)d/(R_2) </span>
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