help me from 8-13 please Find the mean of the data summarized in the given frequ
ID: 3071259 • Letter: H
Question
help me from 8-13 please
Find the mean of the data summarized in the given frequency distribution. 8) The manager of a bank recorded the amount of time each dustomer spent waiting in line 8) during peak business hours one Monday. The frequency distríbution below summarizes the results. Find the mean waiting time Round your answer to one decimal place Waiting time Number of (minutes) customers 4 -7 8-11 12 15 16-19 20-23 24 27 14 Find the median for the given sample data. 9) The weights (in ounces) of 21 cookies are shown. Find the median weight. 0.73 1.06 0.70 1.62 0.71 0.67 1.19 1.06 1.53 0.93 0.73 1.20 1.31 0.62 0.47 1.20 0.67 1.31 1.72 0.71 056 10) The normal monthly precipitation (in inches) for August is listed for 20 different U.S. cities. 10) Find the median of the data. 3.5 1.6 2.4 37 4.1 3.9 1.0 3.6 42 3.4 3.7 2.2 1.5 42 3.4 2.7 0.4 3.7 20 3.6 Find the modets) for the given sample data. 11) The weights (in ounces) of 14 different apples are shown below 6.8 59 5.7 6.5 5.4 6.8 5.9 49 4.8 65 6.8 49 65 4 Find the midrange for the given sample data. 12) The weights (in ounces) of 18 cookies are shown. Find the midrange. 12) 0.60 1.34 0.89 0.96 0.80 1.43 1.34 1.16 0.60 1.49 1.34 1.16 1.34 1.49 0.80 1.34 0.96 0.89 ind the mean, median, mode, and midrange for each of the two samples, then compare the two sets of results. 13) A comparison is made between summer electric bills of those who have central air and 13) those who have window units May June July Aug Sept Central $32 $64 $80 $90 $65 Window $15 $84 $99 $120 $40Explanation / Answer
8) Solution : => Answer : 9.2
Given that waiting time median(x) Number of customers F(x)
0 - 3 1.5 15
4 - 7 5.5 11
8 - 11 9.5 14
12 - 15 13.5 7
16 - 19 17.5 8
20 - 23 21.5 3
24 - 27 25.5 1
=> Mean = sum of x*F(x)/N
= (1.5*15 + 5.5*11 + 9.5*14 + 13.5*7 + 17.5*8 + 21.5*3 + 25.5*1)/59
= 540.5/59
= 9.2
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9) Solution : => Answer : 0.93
Given that 0.73,1.06,0.70,1.62,0.71,0.67,1.19,1.06,1.53,0.93,0.73,1.20,1.31,0.62,0.47,1.20,0.67,1.31,1.72,0.71,0.56
=> The median of the data set is 0.93
Explanation :
=> 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:
0.47 0.56 0.62 0.67 0.67 0.70 0.71 0.71 0.73 0.73 0.93 1.06 1.06 1.19 1.20 1.20 1.31 1.31 1.53 1.62 1.72
=> So, the median is 0.93 .
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10) Solution : => Answer : 3.45
Given that 3.5,1.6,2.4,3.7,4.1,3.9,1.0,3.6,4.2,3.4,3.7,2.2,1.5,4.2,3.4,2.7,0.4,3.7,2.0,3.6
=> The median of the data set is 3.45
Explanation :-
=> 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:
0.4 1.0 1.5 1.6 2.0 2.2 2.4 2.7 3.4 3.4 3.5 3.6 3.6 3.7 3.7 3.7 3.9 4.1 4.2 4.2
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:
=> Median = (3.4 + 3.5)/2 = 3.45
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11) Solution : => Answer : 6.5 and 6.8
Given that 6.8,5.9,5.7,6.5,5.4,6.8,5.9,4.9,4.8,6.5,6.8,4.9,6.5,4.7
=> The modes of the data set are 6.5 and 6.8
Explanation :-
=> The mode of a set of data is the value in the set that occurs most often.
=> Ordering the data from least to greatest, we get:
4.7 4.8 4.9 4.9 5.4 5.7 5.9 5.9 6.5 6.5 6.5 6.8 6.8 6.8
=> Since both 6.5 and 6.8 occur 3 times, the modes are 6.5 and 6.8 (this data set is bimodal).
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12) Solution : => Answer : 1.045
Given that 0.60,1.34,0.89,0.96,0.80,1.43,1.34,1.16,0.60,1.49,1.34,1.16,1.34,1.49,0.80,1.34,0.96,0.89
=> Midrange of the data set is 1.045
Explanation :-
=> Midrange = (high value + low value)/2
=> Ordering the data from least to greatest, we get:
0.60 0.60 0.80 0.80 0.89 0.89 0.96 0.96 1.16 1.16 1.34 1.34 1.34 1.34 1.34 1.43 1.49 1.49
=> The lowest value is 0.60.
=> The highest value is 1.49.
=> Midrange = (0.60 + 1.49)/2 = 1.045
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13) solution :
Given that central air as 32,64,80,90,65
window air as 15,84,99,120,40
=> central air : Mean = 66.2 , Median = 65 , there is no mode for this data, Midrange = 61
=> Window unit : Mean = 71.6 , Median = 84 , there is no mode for this data, Midrange = 67.5
=> Window units appear to be significantly more expensive.
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