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
To approximate the mean and standard deviation for temperature, we need to calculate the midpoint of each temperature range and use it as the representative value for that range. Then, we can use these representative values to calculate the mean and standard deviation.
Here are the midpoints for each temperature range:
50-59: 54.5
60-69: 64.5
70-79: 74.5
80-89: 84.5
90-99: 94.5
100-109: 104.5
Now, we can calculate the mean (μ) by multiplying each midpoint by its corresponding frequency (number of days) and summing them up, then dividing by the total number of days.
μ = (54.5*1 + 64.5*309 + 74.5*1433 + 84.5*1521 + 94.5*412 + 104.5*14) / (1 + 309 + 1433 + 1521 + 412 + 14)
Calculating this, the approximate mean temperature is:
μ ≈ (54.5 + 19984.5 + 1062685.5 + 1285744.5 + 39234 + 1463) / 4730
≈ 3387406 / 4730
≈ 715.57
Therefore, the approximate mean temperature is 715.57.
To calculate the standard deviation, we need to calculate the deviations of each midpoint from the mean, square them, multiply by their corresponding frequency, sum them up, divide by the total number of days, and take the square root of the result.
σ = √(( (54.5-715.57)^2 * 1 + (64.5-715.57)^2 * 309 + (74.5-715.57)^2 * 1433 + (84.5-715.57)^2 * 1521 + (94.5-715.57)^2 * 412 + (104.5-715.57)^2 * 14 ) / (1 + 309 + 1433 + 1521 + 412 + 14))
Calculating this, the approximate standard deviation for temperature is:
σ ≈ √(( ( (-661.07)^2 * 1 + (-651.07)^2 * 309 + (-641.07)^2 * 1433 + (-631.07)^2 * 1521 + (-621.07)^2 * 412 + (-611.07)^2 * 14 ) / 4730))
≈ √((437266404.49 + 42602531314.09 + 738119662126.09 + 816317362486.81 + 47308762.09 + 1110798.06) / 4730)
≈ √((1554216429983.63) / 4730)
≈ √(328420637.87)
≈ 18117.61
Therefore, the approximate standard deviation for temperature is 18117.61.
To approximate the mean and standard deviation for temperature, follow these steps:
Step 1: Assign values to the midpoints of each temperature range:
Midpoint of the first range (50-59) = 55
Midpoint of the second range (60-69) = 65
Midpoint of the third range (70-79) = 75
Midpoint of the fourth range (80-89) = 85
Midpoint of the fifth range (90-99) = 95
Midpoint of the sixth range (100-109) = 105
Step 2: Multiply each midpoint by its corresponding frequency (number of days) to calculate the weighted sum.
Weighted sum = (55 * 1) + (65 * 309) + (75 * 1433) + (85 * 1521) + (95 * 412) + (105 * 14)
Step 3: Calculate the total frequency by summing up all the frequencies.
Total frequency = 1 + 309 + 1433 + 1521 + 412 + 14
Step 4: Divide the weighted sum by the total frequency to find the approximate mean (μ).
μ = weighted sum / total frequency
Step 5: Calculate the sum of squared differences between each midpoint and the mean, multiplied by their corresponding frequencies.
Sum of squared differences = [(55 - μ)^2 * 1] + [(65 - μ)^2 * 309] + [(75 - μ)^2 * 1433] + [(85 - μ)^2 * 1521] + [(95 - μ)^2 * 412] + [(105 - μ)^2 * 14]
Step 6: Calculate the mean of squared differences by dividing the sum of squared differences by the total frequency.
Mean of squared differences = sum of squared differences / total frequency
Step 7: Take the square root of the mean of squared differences to find the approximate standard deviation.
Standard deviation = √(mean of squared differences)
By following these steps and plugging in the given values, you should be able to calculate the approximate mean and standard deviation for the temperature distribution.
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.