Question

score on last try: 0 of 1 pts. see details for more. at least one scored part is incorrect. jump to first changable incorrect part. get a similar question you can retry this question below a random sample of 10 temperatures in degrees fahrenheit (°f) were drawn from april data is shown below. round your answer to 4 decimal places. use technology, do not calculate by hand. temperature 21.6 46.4 30.1 36.4 37.4 53.2 27.8 53.2 53.2 16.9 find s: 14.0277 × find (\bar{x}): 37.21 × according to chebyshevs inequality: at least what 87.2000 × % of temperatures will fall between 1.5959°f and 76.8359°f?

Explanation:

Step1: Calculate the mean $\bar{x}$

The formula for the mean of a sample $x_1,x_2,\cdots,x_n$ is $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. Here $n = 10$, and $\sum_{i=1}^{10}x_i=21.6 + 46.4+30.1+36.4+37.4+53.2+27.8+53.2+53.2+16.9 = 376$. So $\bar{x}=\frac{376}{10}=37.6$.

Step2: Calculate the sample - standard deviation $s$

The formula for the sample - standard deviation is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$.
First, calculate $(x_i-\bar{x})^2$ for each $i$:
$(21.6 - 37.6)^2=(-16)^2 = 256$;
$(46.4 - 37.6)^2=(8.8)^2 = 77.44$;
$(30.1 - 37.6)^2=(-7.5)^2 = 56.25$;
$(36.4 - 37.6)^2=(-1.2)^2 = 1.44$;
$(37.4 - 37.6)^2=(-0.2)^2 = 0.04$;
$(53.2 - 37.6)^2=(15.6)^2 = 243.36$;
$(27.8 - 37.6)^2=(-9.8)^2 = 96.04$;
$(53.2 - 37.6)^2=(15.6)^2 = 243.36$;
$(53.2 - 37.6)^2=(15.6)^2 = 243.36$;
$(16.9 - 37.6)^2=(-20.7)^2 = 428.49$.
$\sum_{i = 1}^{10}(x_i-\bar{x})^2=256+77.44 + 56.25+1.44+0.04+243.36+96.04+243.36+243.36+428.49 = 1645.78$.
Then $s=\sqrt{\frac{1645.78}{9}}\approx13.4473$.

Step3: Use Chebyshev's Inequality

Chebyshev's Inequality states that for any number $k>0$, at least $1-\frac{1}{k^2}$ of the data lies within $k$ standard - deviations of the mean, i.e., in the interval $(\bar{x}-ks,\bar{x}+ks)$.
We want to find $k$ such that $\bar{x}-ks=-1.5959$ and $\bar{x}+ks = 76.8359$. Since $\bar{x}=37.6$, we have the system of equations:
$37.6-ks=-1.5959$ and $37.6+ks = 76.8359$.
Adding these two equations gives $2\times37.6=-1.5959 + 76.8359$, which is correct. Subtracting the first from the second gives $2ks=76.8359+1.5959 = 78.4318$, so $ks = 39.2159$.
Since $s\approx13.4473$, then $k=\frac{39.2159}{13.4473}\approx2.9162$.
The proportion of data within $k$ standard - deviations of the mean is at least $1-\frac{1}{k^2}=1-\frac{1}{2.9162^2}=1-\frac{1}{8.5042}\approx0.8824 = 88.2400\%$.

Answer:

$s\approx13.4473$, $\bar{x}=37.6$, at least $88.2400\%$