Answer:
12
Step-by-step explanation:
When the numerators are the same, the denominator that is bigger has a lesser value. In this case, the inequality > shows that is bigger than the fraction next to it so in order to find a fraction that is lesser in value than
, we need to find a denominator that is <u>more than 10</u>, since 10 is the denominator of the first fraction. The only number more than 10 in the options is 12, so 12 is the answer.
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.
To know more about standard deviation visit:
#SPJ4
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