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 4
60-69 60 69 316
70-79 70 79 1426
80-89 80 89 1532
90-99 90 99 345
100-109 100 109 7

Answer 1

Answer:

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

μ =79.7

б =7.81

Step-by-step explanation:

Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.

Standard deviation is the square root of the arithmetic mean of the squares of deviations of the items from their mean values.

Frequency Distribution

Given that,

Temperature       Lower Limit           Upper Limit             Days($f_{i}$)

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($x_{i\\}$)    

$\frac{50+59}{2} = 54.5\\ \\\frac{60+69}{2} = 64.5\\ \\\frac{70+79}{2} = 74.5\\\\\frac{80+89}{2} = 84.5\\ \\\frac{90+99}{2} = 94.5\\ \\\frac{100+109}{2} = 104.5$

As we know,

Mean (μ) =∑ $(x_{i}f_{i})$/∑$f_{i}$

= { (54.5 * 2) + (64.5 * 313) +(74.5 * 1419)+(84.5*1503)+(94.5*319) + (104.5*9)/ (2+313+1419+1503+319)

∵ μ = 79.7

As we know,

Standard Deviation:-

$\sigma={\sqrt {\frac {\sum(x_{i}-{\mu})^{2}}{N}}}$

by putting the values, we get

∵ б = 7.81

Hence, μ = 79.7 and б=7.81  are the correct answers.

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