Practise special statistics problems involving combined mean, combined standard deviation, weighted mean and corrected mean with corrected standard deviation.
Combined mean is used when two or more groups are joined together. The group sizes must be considered.
Coefficient of variation is useful for comparing variability when means are different.
In correction questions, first recover \(\sum x\) and \(\sum x^2\) from the given mean and standard deviation. Then remove or replace the wrong value.
\[ \bar{x}=\frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2} \]
\[ \sigma=\sqrt{\frac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}} \]
\[ CV=\frac{\text{S.D.}}{\text{Mean}}\times100 \]
\[ \sigma=\sqrt{\frac{\sum x^2}{n}-\bar{x}^{\,2}} \]
In two companies A and B engaged in similar type of industry, the average weekly wage and standard deviation are given below. Find which company pays a larger total amount as weekly wages, which company shows greater variability, and find the mean and standard deviation of all workers taken together.
| Company A | Company B | |
|---|---|---|
| Average Weekly Wage (Rs) | 460 | 490 |
| Standard Deviation | 50 | 40 |
| No. of wage earners | 100 | 80 |
Compare total weekly wages, compare CV, then calculate combined mean and combined SD.
Company A total weekly wages:
\[ 100\times460=46000 \]
Company B total weekly wages:
\[ 80\times490=39200 \]
Therefore, Company A pays a larger total amount as weekly wages.
To compare variability, use coefficient of variation:
\[ CV=\frac{\text{S.D.}}{\text{Mean}}\times100 \]
\[ CV_A=\frac{50}{460}\times100=10.87\% \]
\[ CV_B=\frac{40}{490}\times100=8.16\% \]
Since Company A has the larger CV, Company A shows greater variability.
Combined mean:
\[ \bar{x}=\frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2} \]
\[ \bar{x}=\frac{100(460)+80(490)}{100+80} \]
\[ \bar{x}=\frac{85200}{180}=473.33 \]
For combined standard deviation:
\[ d_1=460-473.33=-13.33,\qquad d_2=490-473.33=16.67 \]
\[ \sigma=\sqrt{\frac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}} \]
\[ \sigma=\sqrt{\frac{100(50^2+(-13.33)^2)+80(40^2+16.67^2)}{180}} \]
\[ \sigma=\sqrt{2322.22} \]
\[ \sigma=48.19 \]
The mean weight of 100 students in a certain class is 59 kg. The mean weight of boys is 65 kg and that of girls is 50 kg. Find the number of boys and girls in the class.
Let \(x\) be the number of boys and \(y\) be the number of girls. Then form two equations.
Let:
\[ x=\text{number of boys} \]
\[ y=\text{number of girls} \]
Since there are 100 students,
\[ x+y=100 \]
Using the given mean weight:
\[ \bar{x}=\frac{\sum fx}{\sum f} \]
\[ 59=\frac{65x+50y}{100} \]
\[ 5900=65x+50y \]
Now solve the two equations:
\[ x+y=100 \]
\[ 65x+50y=5900 \]
From \(x+y=100\),
\[ y=100-x \]
Substitute in \(65x+50y=5900\):
\[ 65x+50(100-x)=5900 \]
\[ 65x+5000-50x=5900 \]
\[ 15x=900 \]
\[ x=60 \]
\[ y=100-60=40 \]
The mean and S.D. of 20 items is found to be 10 and 2 respectively. At the time of checking it was found that one item 8 was incorrect. Calculate the mean and standard deviation if: (a) the wrong item is omitted, and (b) it is replaced by 12.
First recover \(\sum x\) and \(\sum x^2\), then adjust them for omission or replacement.
Given:
\[ n=20,\qquad \bar{x}=10,\qquad \sigma=2 \]
First find \(\sum x\):
\[ \sum x=n\bar{x}=20(10)=200 \]
Using:
\[ \sigma=\sqrt{\frac{\sum x^2}{n}-\bar{x}^{\,2}} \]
\[ 2=\sqrt{\frac{\sum x^2}{20}-10^2} \]
\[ 4=\frac{\sum x^2}{20}-100 \]
\[ \frac{\sum x^2}{20}=104 \]
\[ \sum x^2=2080 \]
(a) If the wrong item \(8\) is omitted:
\[ n'=19 \]
\[ \sum x'=200-8=192 \]
\[ \sum x'^2=2080-8^2=2080-64=2016 \]
\[ \bar{x}'=\frac{192}{19}=10.11 \]
\[ \sigma'=\sqrt{\frac{2016}{19}-(10.11)^2} \]
\[ \sigma'=2.00 \]
(b) If the wrong item \(8\) is replaced by \(12\):
\[ n'=20 \]
\[ \sum x'=200-8+12=204 \]
\[ \sum x'^2=2080-8^2+12^2 \]
\[ \sum x'^2=2080-64+144=2160 \]
\[ \bar{x}'=\frac{204}{20}=10.20 \]
\[ \sigma'=\sqrt{\frac{2160}{20}-(10.20)^2} \]
\[ \sigma'=1.99 \]