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A function 
is periodic if there is some constant 
such that 
for all 
in the domain of . Then is the "period" of .
Example:
If
, then we have 
, and so 
is periodic with period 
.
It gets a bit more complicated for a function like yours. We're looking for such that
Expanding on the left, you have
and
It follows that the following must be satisfied:
The first two equations are satisfied whenever
, or more generally, when 
and 
(i.e. any multiple of 4).
The second two are satisfied whenever
, and more generally when 
with (any multiple of 10/7).
It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.
Let's verify:
More generally, it can be shown that
is periodic with period
.
Example:
If
It gets a bit more complicated for a function like yours. We're looking for
Expanding on the left, you have
and
It follows that the following must be satisfied:
The first two equations are satisfied whenever
The second two are satisfied whenever
It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when
Let's verify:
More generally, it can be shown that
is periodic with period
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