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The following data represent the​ high-temperature distribution for a summer month in a city for some of the last 130 years. Treat the data as a population. Complete parts​ (a) through​ (c).Temperature Lower Limit Upper Limit Days50-59                50                    59            260-69                60                    69          31370-79                 70                    79         141980-89                80                    89         150390-99                90                    99          319100-109            100                  109            9 ​(a) Approximate the mean and standard deviation for temperature.

Question
The following data represent the​ high-temperature distribution for a summer month in a city for some of the last 130 years. Treat the data as a population. Complete parts​ (a) through​ (c).
Temperature Lower Limit Upper Limit Days
50-59                50                    59            2
60-69                60                    69          313
70-79                 70                    79         1419
80-89                80                    89         1503
90-99                90                    99          319
100-109            100                  109            9
 
​(a) Approximate the mean and standard deviation for temperature.
check_circleAnswer
Step 1

Here, first lower case and upper case limits are 50 and 59. Then, the midpoint of the class is 54.5 (=(50+59)/2); so x1 is 54.5. Similarly, calculate the midpoints for the remaining classes.

Step 2

The values of ∑fi, ∑fixi and ∑fixi2 are calculated as follows:

Class Midpoint
Frequency f
Temperature
54.5
50-59
2
109
5,940.5
60-69
64.5
313
20,188.5
1,302,158
74.5
105,715.5
70-79
1,419
7,875,805
84.5
80-89
1,503
127,003.5
10,731,796
90-99
94.5
319
30,145.5
2,848,750
98,282.25
22,862,731
100-109
104.5
9
940.5
Total
477
N-3,565
284,102.5
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Class Midpoint Frequency f Temperature 54.5 50-59 2 109 5,940.5 60-69 64.5 313 20,188.5 1,302,158 74.5 105,715.5 70-79 1,419 7,875,805 84.5 80-89 1,503 127,003.5 10,731,796 90-99 94.5 319 30,145.5 2,848,750 98,282.25 22,862,731 100-109 104.5 9 940.5 Total 477 N-3,565 284,102.5

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Step 3

The mean for temperature is ...

Mean )=
N
284,102.5
3,565
79.69
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Image Transcriptionclose

Mean )= N 284,102.5 3,565 79.69

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