Question

for the following set of data, find the number of data within 1 population standard deviation of the mean.

data	frequency
19	7
20	12
21	14
26	11
34	7
39	2

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Explanation:

Step1: Calculate the mean $\mu$

First, find the sum of the products of data - values and their frequencies. Let $x_i$ be the data - value and $f_i$ be the frequency. The formula for the mean of a frequency - distribution is $\mu=\frac{\sum_{i = 1}^{n}x_if_i}{\sum_{i = 1}^{n}f_i}$.
$\sum_{i = 1}^{n}x_if_i=18\times1 + 19\times7+20\times12 + 21\times14+26\times11+34\times7+39\times2$
$=18+133+240+294+286+238+78$
$=1287$.
$\sum_{i = 1}^{n}f_i=1 + 7+12+14+11+7+2=54$.
So, $\mu=\frac{1287}{54}\approx23.83$.

Step2: Calculate the population standard deviation $\sigma$

The formula for the population standard deviation of a frequency - distribution is $\sigma=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\mu)^2}{\sum_{i = 1}^{n}f_i}}$.
$(18 - 23.83)^2\times1+(19 - 23.83)^2\times7+(20 - 23.83)^2\times12+(21 - 23.83)^2\times14+(26 - 23.83)^2\times11+(34 - 23.83)^2\times7+(39 - 23.83)^2\times2$
$=(- 5.83)^2\times1+(-4.83)^2\times7+(-3.83)^2\times12+(-2.83)^2\times14+(2.17)^2\times11+(10.17)^2\times7+(15.17)^2\times2$
$=34.0+163.4+175.0+113.3+51.6+724.4+460.2$
$=1721.9$.
$\sigma=\sqrt{\frac{1721.9}{54}}\approx5.66$.

Step3: Find the range within 1 standard deviation of the mean

The range is $\mu-\sigma$ to $\mu+\sigma$, so $23.83 - 5.66=18.17$ and $23.83 + 5.66=29.49$.

Step4: Count the number of data within the range

For $x = 18$, frequency $f = 1$ (not in the range).
For $x = 19$, frequency $f = 7$ (in the range).
For $x = 20$, frequency $f = 12$ (in the range).
For $x = 21$, frequency $f = 14$ (in the range).
For $x = 26$, frequency $f = 11$ (in the range).
The total number of data within 1 population standard deviation of the mean is $7 + 12+14+11=44$.

Answer:

44