Respuesta :
Given:
The data and frequency distribution.
Data Frequency
6 2
12 8
26 11
29 12
38 12
45 8
48 2
[tex]\begin{gathered} \operatorname{mean}=\frac{1}{\sum^{}_{}p(x)}\sum ^{}_{}xp(x) \\ =\frac{(6(2)+12(8)+26(11)+29(12)+38(12)+45(8)+48(2))}{2+8+11+12+12+8+2} \\ =\frac{1654}{55} \\ =30.0727 \end{gathered}[/tex]Standard deviation is,
[tex]\begin{gathered} \sigma^2=\frac{\sum^{}_{}(X-\operatorname{mean})^2p(x)}{\sum^{}_{}f(x)} \\ =\frac{(6-30.07)^22+(12-30.07)^28+(26-30.07)^211+(29-30.07)^212+(38-30.07)^212+(45-30.07)^28+(48-30.07)^22}{55} \\ =\frac{1158.7298+2612.1922+182.2139+13.7388+754.6188+1783.2392+642.9698}{55} \\ =129.9582 \\ S\mathrm{}D\mathrm{}=\sqrt[]{\sigma^2} \\ =\sqrt[]{129.9582} \\ =11.3999 \\ \approx11.4 \end{gathered}[/tex]Range of 2 standard deviation is from 30.07-11.4(2) to 30.07+11.4(2)
i.e., 7.27 to 52.87
Count the data point fall in this range that is from 12 to 48 the number of data points are,
8+11+12+12+8+2=53 data points
That means 53% of all data
