angie is researching how pollution might affect the oxygen level in the water of two ponds. she measured the…

angie is researching how pollution might affect the oxygen level in the water of two ponds. she measured the oxygen level of each pond, in parts per million, every week for 5 weeks.\nbarnyard pond\n5 8 5 4 8\nforest pond\n7 6 5 8 9\nwhich pond has a greater standard deviation in its oxygen level?\nbarnyard pond forest pond\nwhat is the standard deviation for that pond? write your answer as a decimal number rounded to the nearest hundredth.\nparts per million

angie is researching how pollution might affect the oxygen level in the water of two ponds. she measured the oxygen level of each pond, in parts per million, every week for 5 weeks.\nbarnyard pond\n5 8 5 4 8\nforest pond\n7 6 5 8 9\nwhich pond has a greater standard deviation in its oxygen level?\nbarnyard pond forest pond\nwhat is the standard deviation for that pond? write your answer as a decimal number rounded to the nearest hundredth.\nparts per million

Answer

Explanation:

Step1: Calculate mean of Barnyard pond

$\bar{x}_{1}=\frac{5 + 8+5+4+8}{5}=\frac{30}{5}=6$

Step2: Calculate squared - differences for Barnyard pond

$(5 - 6)^2=1$, $(8 - 6)^2 = 4$, $(5 - 6)^2=1$, $(4 - 6)^2 = 4$, $(8 - 6)^2=4$ Sum of squared - differences $S_1=1 + 4+1+4+4 = 14$ Variance $s_{1}^{2}=\frac{S_1}{n - 1}=\frac{14}{4}=3.5$ Standard deviation $s_1=\sqrt{3.5}\approx1.87$

Step3: Calculate mean of Forest pond

$\bar{x}_{2}=\frac{7 + 6+5+8+9}{5}=\frac{35}{5}=7$

Step4: Calculate squared - differences for Forest pond

$(7 - 7)^2=0$, $(6 - 7)^2 = 1$, $(5 - 7)^2=4$, $(8 - 7)^2 = 1$, $(9 - 7)^2=4$ Sum of squared - differences $S_2=0 + 1+4+1+4 = 10$ Variance $s_{2}^{2}=\frac{S_2}{n - 1}=\frac{10}{4}=2.5$ Standard deviation $s_2=\sqrt{2.5}\approx1.58$

Step5: Compare standard deviations

Since $1.87>1.58$, Barnyard pond has a greater standard deviation.

Answer:

Barnyard pond 1.87