Question

a company had 80 employees whose salaries are summarized in the frequency distribution below. find the standard deviation using the following formula, where x represents the class mid - point, f represents the class frequency, and n represents the total number of sample values. s = \sqrt{\frac{n\sum(f\cdot x^{2})-\sum(f\cdot x)^{2}}{n(n - 1)}} salary (dollars) employees 5,001 - 10,000 19 10,001 - 15,000 14 15,001 - 20,000 12 20,001 - 25,000 16 25,001 - 30,000 19 a. $8195.1 b. $8422.8 c. $7967.5 d. $7588.1

Explanation:

Step1: Calculate class mid - points

For 5001 - 10000, $x_1=\frac{5001 + 10000}{2}=7500.5$; for 10001 - 15000, $x_2=\frac{10001+15000}{2}=12500.5$; for 15001 - 20000, $x_3=\frac{15001 + 20000}{2}=17500.5$; for 20001 - 25000, $x_4=\frac{20001+25000}{2}=22500.5$; for 25001 - 30000, $x_5=\frac{25001+30000}{2}=27500.5$.

Step2: Calculate $f\cdot x$ and $f\cdot x^{2}$ for each class

For the first class: $f_1 = 19$, $f_1x_1=19\times7500.5 = 142509.5$, $f_1x_1^{2}=19\times(7500.5)^{2}=19\times56257500.25 = 1068892504.75$.
For the second class: $f_2 = 14$, $f_2x_2=14\times12500.5 = 175007$, $f_2x_2^{2}=14\times(12500.5)^{2}=14\times156262500.25 = 2187675003.5$.
For the third class: $f_3 = 12$, $f_3x_3=12\times17500.5 = 210006$, $f_3x_3^{2}=12\times(17500.5)^{2}=12\times306262500.25 = 3675150003$.
For the fourth class: $f_4 = 16$, $f_4x_4=16\times22500.5 = 360008$, $f_4x_4^{2}=16\times(22500.5)^{2}=16\times506272500.25 = 8100360004$.
For the fifth class: $f_5 = 19$, $f_5x_5=19\times27500.5 = 522509.5$, $f_5x_5^{2}=19\times(27500.5)^{2}=19\times756272500.25 = 14369177504.75$.

Step3: Calculate $\sum(f\cdot x)$ and $\sum(f\cdot x^{2})$

$\sum(f\cdot x)=142509.5 + 175007+210006+360008+522509.5=1410040$.
$\sum(f\cdot x^{2})=1068892504.75+2187675003.5+3675150003+8100360004+14369177504.75 = 30021255020$.

Step4: Calculate the standard deviation

$n = 80$.

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The closest answer is A. $8195.1$.

Answer:

A. $8195.1$