SOLUTION: 3) The following frequency distribution reports the electricity cost for a sample of 50 two-bedroom apartments in Jenin, Palestine, during the month of May last year. Electrici

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Question 1162753:
3) The following frequency distribution reports the electricity cost for a sample of 50 two-bedroom apartments in Jenin, Palestine, during the month of May last year.
Electricity cost frequency
80 up to 100 4
100 up to 120 7
120 up to 140 13
140 up to 160 16
160 up to 180 7
180 up to 200 3
Total 50
1. Estimate the mean cost.


2. Estimate the standard deviation.



3. What is the approximated percentage of families in Jenin, Palestine that live in two-bedroom apartments that pay at least 140 and less than 180 shekels during May last year?








Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!

Problem 1

Here is the original table
Electricity Cost (c)Frequency (f)
80 <= c < 1004
100 <= c < 1207
120 <= c < 14013
140 <= c < 16016
160 <= c < 1807
180 <= c < 2003

where c = cost of electricity and f represents the frequency that goes with it. When I write something like 80 <= c < 100, I mean that c is between 80 and 100. Furthermore, c could be 80, but c cannot be 100.

Form a new column of values that are the midpoints of the intervals given. The midpoint between 80 and 100 is 90, since (80+100)/2 = 180/2 = 90. The midpoint of 100 and 120 is 110 for similar reasoning. You could repeatedly apply the midpoint formula, or note that the midpoints 90, 110, ... are going up by 20. This is because the start points and endpoints of each interval also increase by 20. Everything is in lockstep together.

Here's what the table should look like after you have added the midpoint column. I'm making x be the midpoint values
Electricity Cost (c)Frequency (f)midpoint cost (x)
80 <= c < 100490
100 <= c < 1207110
120 <= c < 14013130
140 <= c < 16016150
160 <= c < 1807170
180 <= c < 2003190

The midpoint is a representative of the entire interval. For some interval like 80 <= c < 100, we don't know what the exact value of c is (as c is likely to vary anyway). So the next best thing is to just pick the most representative value from the interval. The center is the best bet.

Next step is to multiply the corresponding x and f values to form a new column x*f
Electricity Cost (c)Frequency (f)midpoint cost (x)x*f
80 <= c < 100490360
100 <= c < 1207110770
120 <= c < 140131301690
140 <= c < 160161502400
160 <= c < 18071701190
180 <= c < 2003190570

For example, in row 2 we have x*f = 110*7 = 770. The other rows are treated the same.

We're almost done. Add up everything in the x*f column
360+770+1690+2400+1190+570 = 6980

Finally, divide that sum by n = 50 because the sum of the frequency (f) values is 50. This will yield the mean
mean = (sum of x*f)/(sum of f)
mean = 6980/50
mean = 139.6

We will use the mean later for problem 2. We'll call the mean m, so m = 139.6

Answer: estimated mean cost is approximately 139.6 shekels

========================================================
Problem 2

Go back to the original table
Electricity Cost (c)Frequency (f)
80 <= c < 1004
100 <= c < 1207
120 <= c < 14013
140 <= c < 16016
160 <= c < 1807
180 <= c < 2003


Subtract the mean (m) from each x value. Then square the result. Lastly, multiply that square by the frequency f value
This will form a new column labeled f*(x-m)^2
Electricity Cost (c)Frequency (f)midpoint cost (x)f*(x-m)^2
80 <= c < 1004909840.64
100 <= c < 12071106133.12
120 <= c < 140131301198.08
140 <= c < 160161501730.56
160 <= c < 18071706469.12
180 <= c < 20031907620.48

For instance, row 3 has the following calculation
f*(x-m)^2 = 13*(130-139.6)^2 = 1198.08

Add up everything in the f*(x-m)^2 column to get
9840.64+6133.12+1198.08+1730.56+6469.12+7620.48 = 32,992

Divide this over n-1 = 50-1 = 49 to get the sample variance
sample variance = 32992/49 = 673.30612244898

Apply the square root to get the sample standard deviation
sqrt( 673.30612244898 ) = 25.9481429479833


Answer: Approximately 25.9481429479833
Round this value however you need to

Side note: we want the sample standard deviation (rather than the population standard deviation) because our sample of n = 50 families represents a larger population in which we don't know about. We only use a population standard deviation if we have data on the entire population (ie a census).

========================================================
Problem 3

Original table
Electricity Cost (c)Frequency (f)
80 <= c < 1004
100 <= c < 1207
120 <= c < 14013
140 <= c < 16016
160 <= c < 1807
180 <= c < 2003



Highlight the rows where the cost c is between 140 and 200
Electricity Cost (c)Frequency (f)
80 <= c < 1004
100 <= c < 1207
120 <= c < 14013
140 <= c < 16016
160 <= c < 1807
180 <= c < 2003

The frequencies here are 16, 7 and 3 which add to 16+7+3 = 26

We have 26 families that fit in this range, out of 50 families total. So approximately 26/50 = 52/100 = 0.52 = 52% of the families pay in the range of 140 to 180 shekels. This is of course an estimate because a sample is an estimate or approximation of the population.

Answer: About 52%
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