Respuesta :
Answer:
[tex]\sigma =0.48 (to 2 significant figures)[/tex]
Step-by-step explanation:
Standard Deviation [tex]\sigma =\sqrt{\frac{\sum(x-\mu)^2}{N}[/tex] where μ=mean, N=number of data
The results are:x= 36.1, 35.7, and 35.8
First, we determine the mean
Mean, μ= (36.1 + 35.7 + 35.8)/3=107.6/3=35.9
Next, for each x, we determine the mean deviation, x-μ
If x=36.1, |x-μ|=|35.1-35.9|=0.8 and [tex]|x-\mu|^{2}[/tex]=0.64
If x=35.7, |x-μ|=|35.7-35.9|=0.2 and [tex]|x-\mu|^{2}[/tex]=0.04
If x=35.8, |x-μ|=|35.8-35.9|=0.1 and [tex]|x-\mu|^{2}[/tex]=0.01
[tex]\sigma =\sqrt{\frac{\sum(x-\mu)^2}{N}}=\sqrt{\frac{0.64+0.04+0.01}{3} } =\sqrt{\frac{0.69}{3} }=0.4796[/tex]
[tex]\sigma =0.48 (to 2 significant figures)[/tex]
the sample standard deviation of these measurements is 0.21
given that three readings of salinity by a group of students are:
36.1
35.7
35.8
what is the sample standard deviation?
The standard deviation is a statistic that measures the dispersion of a dataset relative to its mean and is calculated as the square root of the variance.
as we know the standard deviation formulae,
σ = [Σ(Xₙ - μ)²/(n-1)]^0.5
where n = number of observations where n=1,2,3....
μ= mean of observations.
mean of all three readings
μ= (36.1+35.7+35.8)/3
μ=35.87
X₁= 36.1
X₂=35.7
X₃= 35.8
NOW, standard deviation,
σ = [tex]\sqrt{\frac{(36.1-35.87)^2 + (35.7-35.87)^2+ (35.8-35.87)^2}{2} }[/tex]
σ = 0.21
Hence, the sample standard deviation of these measurements is 0.21
To get more about standard deviation refer to the link,
https://brainly.com/question/12402189