Use the table to answer the question.
Value 4 5 6
Total Observations 1 3 1
Calculate the standard deviation of the data set. Round the answer to the nearest hundredth.
To calculate the standard deviation, we first need to find the mean of the data set.
Mean = (4*1 + 5*3 + 6*1) / (1+3+1) = (4 + 15 + 6) / 5 = 25 / 5 = 5
Next, calculate the squared differences between each value and the mean, and then multiply each squared difference by the frequency of that value and sum them up:
(4-5)^2 * 1 + (5-5)^2 * 3 + (6-5)^2 * 1
= 1 * 1 + 0 * 3 + 1 * 1
= 1 + 0 + 1
= 2
Now, divide the sum by the total number of observations minus 1, and then take the square root of that result to get the standard deviation:
Standard deviation = sqrt(2 / (5-1))
= sqrt(2 / 4)
= sqrt(0.5)
≈ 0.71
Therefore, the standard deviation of the data set is approximately 0.71.
Use the table to answer the question.
Value 3 4 5
Total Observations 2 3 2
Charles records the number of miles he walks every day. The chart includes how many days he walked the given number of miles for one week. Calculate the standard deviation for the number of miles he walked. Round the answer to the nearest hundredth.
To calculate the standard deviation, we first need to find the mean of the data set.
Mean = (3*2 + 4*3 + 5*2) / (2+3+2) = (6 + 12 + 10) / 7 = 28 / 7 = 4
Next, calculate the squared differences between each value and the mean, and then multiply each squared difference by the frequency of that value and sum them up:
(3-4)^2 * 2 + (4-4)^2 * 3 + (5-4)^2 * 2
= 1 * 2 + 0 * 3 + 1 * 2
= 2 + 0 + 2
= 4
Now, divide the sum by the total number of observations minus 1, and then take the square root of that result to get the standard deviation:
Standard deviation = sqrt(4 / (7-1))
= sqrt(4 / 6)
= sqrt(0.67)
≈ 0.82
Therefore, the standard deviation for the number of miles Charles walked is approximately 0.82.
Compare and interpret the measures to select the data with the least variation.(1 point)
Responses

Dataset A has a standard deviation of 1.9 and a mean of 21.
Dataset A has a standard deviation of 1.9 and a mean of 21.

Dataset B has a standard deviation of 1.41 and a mean of 29.
Dataset B has a standard deviation of 1.41 and a mean of 29.

Dataset C has a standard deviation of 1.2 and a mean of 25.
Dataset C has a standard deviation of 1.2 and a mean of 25.

Dataset D has a standard deviation of 1.14 and a mean of 48.
To select the data with the least variation, we need to compare the standard deviations of the datasets. The dataset with the smallest standard deviation will have the least variation.
Comparing the standard deviations:
Dataset A: 1.9
Dataset B: 1.41
Dataset C: 1.2
Dataset D: 1.14
Dataset D has the smallest standard deviation of 1.14, which means it has the least variation in the data compared to the other datasets. Therefore, Dataset D is the data with the least variation.
If the mean of a dataset is 59, what is the deviation of the data point at 43?
To find the deviation of a specific data point from the mean, you simply subtract the mean from the data point. In this case, to find the deviation of the data point at 43 from the mean of 59:
Deviation = 43 - 59 = -16
Therefore, the deviation of the data point at 43 from the mean of 59 is -16.
Use the standard deviation to compare the variability of the datasets. Round to the nearest hundredths.
Dataset A: 11, 7, 12, 8, 9, 5
Dataset B: 10, 8, 10, 8, 8, 4
To compare the variability of the datasets, we will calculate the standard deviation of each dataset and then compare them by rounding to the nearest hundredth.
**Dataset A:**
Mean = (11 + 7 + 12 + 8 + 9 + 5) / 6 = 52 / 6 = 8.67
Calculate the squared differences:
(11-8.67)^2 = 7.84
(7-8.67)^2 = 6.25
(12-8.67)^2 = 11.71
(8-8.67)^2 = 0.45
(9-8.67)^2 = 0.11
(5-8.67)^2 = 13.28
Sum of squared differences = 39.64
Standard deviation for Dataset A:
sqrt(39.64 / 6) ≈ sqrt(6.607) ≈ 2.57
**Dataset B:**
Mean = (10 + 8 + 10 + 8 + 8 + 4) / 6 = 48 / 6 = 8
Calculate the squared differences:
(10-8)^2 = 4
(8-8)^2 = 0
(10-8)^2 = 4
(8-8)^2 = 0
(8-8)^2 = 0
(4-8)^2 = 16
Sum of squared differences = 24
Standard deviation for Dataset B:
sqrt(24 / 6) = sqrt(4) = 2
Therefore, the standard deviation for Dataset A is approximately 2.57, and for Dataset B is 2. The larger standard deviation in Dataset A indicates that it has more variability compared to Dataset B.