Question
Question Use the table to answer the question. Minutes the Bus Came Late −1 0 1 2 3 Number of Days 1 2 4 2 1 Addie’s bus pickup time is 8:00. The table shows the minutes before and after 8:00 the bus arrived at the stop for the last 10 days. Negative values represent minutes before 8:00. Calculate the standard deviation for the number of minutes the bus arrived before or after 8:00. Round your answer to the nearest tenth. (1 point) The standard deviation is .
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GPT 3.5
1.51 minutes.
To calculate the standard deviation, first calculate the mean. The total number of days is 10, so the sum of the number of minutes is 1*(-1) + 2*0 + 4*1 + 2*2 + 1*3 = 1 + 0 + 4 + 4 + 3 = 12.
The mean is then 12 / 10 = 1.2 minutes.
Next, calculate the squared differences between each data point and the mean, then find the average of these squared differences (variance).
(-1-1.2)^2 = 2.44
(0-1.2)^2 = 1.44
(1-1.2)^2 = 0.04
(2-1.2)^2 = 0.64
(3-1.2)^2 = 2.44
(2.44 + 1.44 + 0.04 + 0.64 + 2.44) / 10 = 6.0 / 10 = 0.6
The standard deviation is the square root of the variance, so √0.6 ≈ 1.51 minutes.
To calculate the standard deviation, first calculate the mean. The total number of days is 10, so the sum of the number of minutes is 1*(-1) + 2*0 + 4*1 + 2*2 + 1*3 = 1 + 0 + 4 + 4 + 3 = 12.
The mean is then 12 / 10 = 1.2 minutes.
Next, calculate the squared differences between each data point and the mean, then find the average of these squared differences (variance).
(-1-1.2)^2 = 2.44
(0-1.2)^2 = 1.44
(1-1.2)^2 = 0.04
(2-1.2)^2 = 0.64
(3-1.2)^2 = 2.44
(2.44 + 1.44 + 0.04 + 0.64 + 2.44) / 10 = 6.0 / 10 = 0.6
The standard deviation is the square root of the variance, so √0.6 ≈ 1.51 minutes.