Asked by Nakita
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.
TemperatureTemperature
50-59
60-69
70-79
80-89
90-99
100-109
DaysDays
1
309
1433
1521
412
14
(a) Approximate the mean and standard deviation for temperature.
muμequals=
nothing
TemperatureTemperature
50-59
60-69
70-79
80-89
90-99
100-109
DaysDays
1
309
1433
1521
412
14
(a) Approximate the mean and standard deviation for temperature.
muμequals=
nothing
Answers
Answered by
PsyDAG
Find the mean first = sum of scores/number of scores
55*1 + 65*309 + 75*1433....
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
I'll let you do the calculations.
55*1 + 65*309 + 75*1433....
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
I'll let you do the calculations.
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