Mathematics, 16.10.2020 14:01 kadariusmerchan

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.

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Answer from: stoog7352

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

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

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

=> $\sigma =7.81$

Answer from: Quest

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

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