The mean and (sample) standard deviation σ = 0.2098.
<h3>What exactly would the standard deviation indicate?</h3>
The term "standard deviation" (or "") refers to the degree of dispersion of the data from the mean. Data are grouped around the mean when the standard deviation is low, and are more dispersed when the standard deviation is high.
<h3>According to given information:</h3>
The mean is the product of the dataset's total and the sample size. Mathematically.
The individual periods are Xi.
The sample size is N.
= 1.06 + 1.31 + 1.28 + 0.99,+ 1.48 + 1.37+ 0.98 + 1.31 + 1.59 + 1.55
= 12.92
N = 10
While substituting the value we get:
x = 12.96/10
x = 1.292
The samples' average is 1.292.
The standard deviation:
= (1.48-1.292)^2+(1.37-1.292)^2+(0.98-1.292)^2+(1.31-1.292)^2+(1.59-1.292)^2+(1.55-1.292)^2.
= 0.43996
Putting into the formula we get:
σ = √(0.043996)
σ = 0.2098
The mean and (sample) standard deviation σ = 0.2098.
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I understand that the question you are looking for is:
You measure the period of a mass oscillating on a vertical spring ten times as follows:
Period (s): 1.06, 1.31, 1.28, 0.99, 1.48, 1.37, 0.98, 1.31, 1.59, 1.55
Required:
What are the mean and (sample) standard deviation?
a. Mean: 1.228, Standard Deviation: 0.2135
b. Mean: 1.325, Standard Deviation: 0.1674
c. Mean: 1.292. Standard Deviation: 0.2211
d. Mean: 1.228, Standard Deviation: 0.2098
e. Mean: 1.292, Standard Deviation: 0.2135
tells us that: When we add 5 to a variable x, we get 20. As it has a unique value for x and is completely equal to it(i.e. 15), It is an equality.
tells us that: A variable x equals to 5. Hence, as x is unique for 5 and is wholly equal to it, it's an equality too.
tells us that: A variable x isn't 5 but lesser than it. As we cannot equate it to 5, nor we are given the nature of the variable x, it is an Inequality.
is an expression; It can't be called an equation or an inequality unless we relate it with another expression.