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

Required:
Approximate the mean and standard deviation for temperature.

Answer:

The mean is  [tex]\= x  = 79.7 [/tex]

The standard deviation  [tex]\sigma =7.81[/tex]

Step-by-step explanation:

From the question we are told that

  The data given is  

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        

Generally the average of each limit is evaluated as

Temperature    Lower Limit   Upper Limit         Average[tex](x_i)[/tex]                    Days[tex](f_i)[/tex]

50-59                 50                   59              ( 50 + 59 )/2 =54.5                 2

60-69                 60                   69              (60+69)/2 = 64.5                    313

70-79                 70                    79               (70+79)/2 =74.5                    1419

80-89                 80                   89                (80+89)/2 =84.5                  1503

90-99                 90                   99                 (90+99)/2 = 94.5                319

100-109             100                  109                 (100+109)/2 = 104.5            9  

Generally the mean is  evaluated as

     [tex]\mu  =  \frac{\sum x_i f_i}{\sum f_i}[/tex]

=>  [tex]\mu  =  \frac{ (54.5 * 2) + (64.5 * 313) +\cdots +(104.5 *  9)}{2 + 313 +\cdots + 9}[/tex]

=> [tex]\= x  = 79.7 [/tex]

Generally the standard deviation  is  evaluated as

    [tex]\sigma = \sqrt{ \frac{\sum[ f_i * ( x_i^2 - \= x^2)] }{\sum f_i}}[/tex]

=>   [tex]\sigma =\sqrt{ \frac{[2 * [(54.5)^2) - 79.7^2 ]]+[313 *[ (64.5)^2)-79.7^2] +\cdots +[9 *[ (104.5)^2)- 79.7^2]  }{2 + 313 +\cdots + 9}}[/tex]

=>  [tex]\sigma =7.81[/tex]

The mean and Standard deviation for the given data would be as follows:

μ [tex]= 79.7[/tex]

б [tex]= 7.81[/tex]

Given that,

Temperature       Lower Limit           Upper Limit             Days([tex]f_{i}[/tex])

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  

Average([tex]x_{i}[/tex])    

[tex]( 50 + 59 )/2 =54.5[/tex]

[tex](60+69)/2 = 64.5[/tex]

[tex](70+79)/2 =74.5[/tex]

[tex](80+89)/2 =84.5[/tex]

[tex](90+99)/2 = 94.5[/tex]

[tex](100+109)/2 = 104.5[/tex]

As we know,

Mean (μ) [tex]=[/tex] ∑[tex](x_{i} f_{i}) /[/tex]∑[tex]f_{i}[/tex]

[tex]= [(54.5[/tex] × [tex]2) + (64.5[/tex] ×[tex]313)+....(104.5[/tex]× [tex]9)][/tex][tex]/(2 + 313 + 1419 + ....9)[/tex]

∵ μ [tex]= 79.7[/tex]

As we know,

Standard Deviation:-

[tex]\sigma={\sqrt {\frac {\sum(x_{i}-{\mu})^{2}}{f_{i} }}}[/tex]

by putting the values, we get

∵ б [tex]= 7.81[/tex]

Thus, μ [tex]= 79.7[/tex] and б [tex]= 7.81[/tex] are the correct answers.

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