Situation 1. A small warehouse employs a supervisor at $1,200 a week. An invento
ID: 3327339 • Letter: S
Question
Situation 1. A small warehouse employs a supervisor at $1,200 a week. An inventory manager at $700 a week, six stock boys at $400 a week, and four drivers at $500 a week. 1. Find the 5 numbers summary statistic for his data. 2. Find the mean (average) wage 3. How many employees earn more than the mean wage? Which measure of center best describe a typical wage at this company: The mean or the median? 4. 5. Find the inter-quartile Range (IQR) IQR = 6. Find the standard Deviation. 7. Find the Range Range = Which measure of spread would best describe the payroll: the range, the 1QR, or the standard deviation? 8.Explanation / Answer
Question 1:
Here the distribution given to us is:
400, 400, 400, 400, 400, 400, 500, 500, 500, 500, 700, 1200
Therefore min value = 400, Max value = 1200 and then as there are 12 people, therefore the median would be computed as the average of the 6th and the 7th value that is the average of 400 and 500. Therefore,
Median = 450
The first quartile wage would clearly be 400 as all the wages less than the median are 400.
Now for 12 values, 0.75*(n+1) = 0.75*13 = 9.75th value. Now as both 9th and 10th values are 500. Therefore the third quartile value would be 500.
Therefore the 5 number summary here would be given as:
Xmin = 400, q1 = 400, q2 = 450, q3 = 500 and Xmax = 1200
Question 2:
The mean age here is computed as:
= Sum of all wages / 12
= ( 1200 + 700 + 6*400 + 4*500 ) / 12
= 6300 / 12
= 525
Therefore the mean wage here is 525
Question 3:
There are only 2 people with wage more than the mean wage of 525
Question 4:
Here there are presence of outliers, therefore the median wage that is 450 represents the best measure of central tendency.
Question 5:
The interquartile range is computed as:
IQR = q3 - q1 = 500 - 400 = 100
Therefore IQR = 100