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luda_lava [24]
3 years ago
8

The following data represent the​ high-temperature distribution for a summer month in a city for some of the last 130 years. Tre

at 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.
Mathematics
1 answer:
geniusboy [140]3 years ago
8 0

Answer:

The mean is  \= x  = 79.7

The standard deviation  \sigma =7.81

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(x_i)                    Days(f_i)

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

     \mu  =  \frac{\sum x_i f_i}{\sum f_i}

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

=> \= x  = 79.7

Generally the standard deviation  is  evaluated as

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

=>   \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}}

=>  \sigma =7.81

   

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