Question

A student group measures the salinity of a water sample three times, obtaining the following results: 36.1, 35.7, and 35.8 ppt. Calculate the sample standard deviation of these measurements. Enter your result in decimal notation with an appropriate number of significant figures. Enter only the numbers, omitting the units.

Answer 1

Answer:

$\sigma =0.48 (to 2 significant figures)$

Step-by-step explanation:

Standard Deviation $\sigma =\sqrt{\frac{\sum(x-\mu)^2}{N}$ 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 $|x-\mu|^{2}$=0.64

If x=35.7, |x-μ|=|35.7-35.9|=0.2 and $|x-\mu|^{2}$=0.04

If x=35.8, |x-μ|=|35.8-35.9|=0.1 and $|x-\mu|^{2}$=0.01

$\sigma =\sqrt{\frac{\sum(x-\mu)^2}{N}}=\sqrt{\frac{0.64+0.04+0.01}{3} } =\sqrt{\frac{0.69}{3} }=0.4796$

$\sigma =0.48 (to 2 significant figures)$

Answer 2

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  = $\sqrt{\frac{(36.1-35.87)^2 + (35.7-35.87)^2+ (35.8-35.87)^2}{2} }$

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