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For the following set of data, find the percentage of data within 2 population standarddeviations of the mean, to the nearest 10th of a percent.Data Frequency62128261129123812458482

User Rodmentou
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1 Answer

28 votes
28 votes

Given:

The data and frequency distribution.

Data Frequency

6 2

12 8

26 11

29 12

38 12

45 8

48 2


\begin{gathered} \operatorname{mean}=\frac{1}{\sum^{}_{}p(x)}\sum ^{}_{}xp(x) \\ =((6(2)+12(8)+26(11)+29(12)+38(12)+45(8)+48(2)))/(2+8+11+12+12+8+2) \\ =(1654)/(55) \\ =30.0727 \end{gathered}

Standard deviation is,


\begin{gathered} \sigma^2=\frac{\sum^{}_{}(X-\operatorname{mean})^2p(x)}{\sum^{}_{}f(x)} \\ =((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) \\ =(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}

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

User Jwodder
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2.9k points