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

Answer: standard deviation= 0.21

Step-by-step explanation:

in the attachment