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for the following set of data, find the population standard deviation, …

Question

for the following set of data, find the population standard deviation, to the nearest thousandth
data frequency
2 8
5 3
8 8
13 8
22 4
23 8
24 3
27 9
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Explanation:

Step1: Calculate the total number of data points ($N$)

To find $N$, we sum up all the frequencies.
The frequencies are 8, 3, 8, 8, 4, 8, 3, 9.
So, $N = 8 + 3 + 8 + 8 + 4 + 8 + 3 + 9 = 51$.

Step2: Calculate the population mean ($\mu$)

The formula for the population mean with frequency is $\mu=\frac{\sum (x\cdot f)}{\sum f}$, where $x$ is the data value and $f$ is the frequency.
First, calculate $\sum (x\cdot f)$:

  • For $x = 2$, $f = 8$: $2\times8 = 16$
  • For $x = 5$, $f = 3$: $5\times3 = 15$
  • For $x = 8$, $f = 8$: $8\times8 = 64$
  • For $x = 13$, $f = 8$: $13\times8 = 104$
  • For $x = 22$, $f = 4$: $22\times4 = 88$
  • For $x = 23$, $f = 8$: $23\times8 = 184$
  • For $x = 24$, $f = 3$: $24\times3 = 72$
  • For $x = 27$, $f = 9$: $27\times9 = 243$

Now sum these up: $16 + 15 + 64 + 104 + 88 + 184 + 72 + 243 = 786$

Then, $\mu=\frac{786}{51}\approx15.41176$

Step3: Calculate the sum of squared deviations multiplied by frequency ($\sum (x - \mu)^2\cdot f$)

For each data point:

  • For $x = 2$, $f = 8$: $(2 - 15.41176)^2\times8 = (-13.41176)^2\times8\approx179.875\times8 = 1439.0$ (approximate, more precise calculation: $(-13.41176)^2 = 179.8753$, $179.8753\times8 = 1439.0024$)
  • For $x = 5$, $f = 3$: $(5 - 15.41176)^2\times3 = (-10.41176)^2\times3\approx108.404\times3 = 325.212$ (precise: $(-10.41176)^2 = 108.404$, $108.404\times3 = 325.212$)
  • For $x = 8$, $f = 8$: $(8 - 15.41176)^2\times8 = (-7.41176)^2\times8\approx54.934\times8 = 439.472$ (precise: $(-7.41176)^2 = 54.934$, $54.934\times8 = 439.472$)
  • For $x = 13$, $f = 8$: $(13 - 15.41176)^2\times8 = (-2.41176)^2\times8\approx5.816\times8 = 46.528$ (precise: $(-2.41176)^2 = 5.816$, $5.816\times8 = 46.528$)
  • For $x = 22$, $f = 4$: $(22 - 15.41176)^2\times4 = (6.58824)^2\times4\approx43.405\times4 = 173.62$ (precise: $(6.58824)^2 = 43.405$, $43.405\times4 = 173.62$)
  • For $x = 23$, $f = 8$: $(23 - 15.41176)^2\times8 = (7.58824)^2\times8\approx57.582\times8 = 460.656$ (precise: $(7.58824)^2 = 57.582$, $57.582\times8 = 460.656$)
  • For $x = 24$, $f = 3$: $(24 - 15.41176)^2\times3 = (8.58824)^2\times3\approx73.758\times3 = 221.274$ (precise: $(8.58824)^2 = 73.758$, $73.758\times3 = 221.274$)
  • For $x = 27$, $f = 9$: $(27 - 15.41176)^2\times9 = (11.58824)^2\times9\approx134.308\times9 = 1208.772$ (precise: $(11.58824)^2 = 134.308$, $134.308\times9 = 1208.772$)

Now sum all these up:
$1439.0024 + 325.212 + 439.472 + 46.528 + 173.62 + 460.656 + 221.274 + 1208.772$
First, add step by step:
$1439.0024 + 325.212 = 1764.2144$
$1764.2144 + 439.472 = 2203.6864$
$2203.6864 + 46.528 = 2250.2144$
$2250.2144 + 173.62 = 2423.8344$
$2423.8344 + 460.656 = 2884.4904$
$2884.4904 + 221.274 = 3105.7644$
$3105.7644 + 1208.772 = 4314.5364$

Step4: Calculate the population standard deviation ($\sigma$)

The formula for population standard deviation is $\sigma=\sqrt{\frac{\sum (x - \mu)^2\cdot f}{N}}$
We have $\sum (x - \mu)^2\cdot f = 4314.5364$ and $N = 51$
So, $\frac{4314.5364}{51}\approx84.59875$
Then, $\sigma=\sqrt{84.59875}\approx9.198$

Answer:

The population standard deviation, to the nearest thousandth, is $\boxed{9.198}$