The Mayl, 2009, issue of the Mont Clarian reported the following home sale amoun
ID: 3052136 • Letter: T
Question
The Mayl, 2009, issue of the Mont Clarian reported the following home sale amounts for a sample of homes in Almeda, CA that were sold the previous month 000s of $): 590, 815, 575, 608, 350, 1285, 408, 540, 555, 679 (a) Calculate and interpret the sample mean and median (b) Suppose the 6th observation has been 985 rather than 1285. How would the mean and the 4. median change? (c) Calculate a trimmed mean by trimming the two smallest observations and the two largest observations 5. The data for blood mercury concentration (ig/g) for adult females near contaminated rivers in Virginia are as following: 20 22 .25 .30 .34 .41 .55 .56 142 1.70 1.83 2.20 2.25 3.07 3.25 (a) Determine the values of the sample mean and sample median and explain why they are different (b) By how much could the observation .20 be increased without impacting the sample median? 6. A sample of n 10 automobiles was selected, and each was subjected to a 5-mph crash test. Denoting a car with no visible damage by S (for success) and a car with such damage by F, results were as follows: S SFSSF SF S F (a) What is the value of the sample proportion of successes x/n? (b) Replace each S with a 1 and each F with a 0. Then calculate x for this numerically coded sample.Explanation / Answer
a)
mean = sum of all terms / number of terms = 6405/10 = 640.5
median
The median is the middle number in a sorted list of numbers. So, to find the median, we need to place the numbers in value order and find the middle number.
Ordering the data from least to greatest, we get:
350 408 540 555 575 590 608 679 815 1285
As you can see, we do not have just one middle number but we have a pair of middle numbers, so the median is the average of these two numbers:
= (575+ 590)/2 = 582.5
b)
350 408 540 555 575 590 608 679 815 985
mean = 6105/10 = 610.5
median remain same
c)
trimming two minimum and two maximum
mean = 3547/6
= 591.1667
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