Statistics

(i) The number of wage earner in firm A, n_A=586. Mean monthly wages of firm

Total no of were canner in firm B.

Total monthly mages in firm. B = ?5253 * n_B

=? 5253 * 586

=? 34, 03, 944

Firm B pays larger amount of monthly wages.

(ii) Since both the firm A and B has same mean monthly wages the firm with greater standard duration i.e, greater variance will have more variability in individual ways. Therefore, firm B will have more variability in individual wages.

24. We can tabulate the given data as follows.

Total =

Mean of $x,\overline{x}=\frac{{\displaystyle \sum _{}^{}xi}}{n}=\frac{510}{10}=51.$

Mean of $Y,\overline{Y}=\frac{{\displaystyle \sum _{}^{}Yi}}{n}=\frac{1050}{10}=105$ .

23. Let the assumed mean be A=45 and h=10. Then we can tabulate the given data as following.

$\begin{array}{c}=\frac{-6}{150}*10+45.\\ \end{array}$

= 44.6

22. The given data is converted into continuous frequency duration by subtracting and adding 0.5 from lower and upper limit respectively. Lit the assumed mean be A=42.5 and h=4

21. Let the assumed mean be A=92.5 and h=5

20.

We have,  $N={\displaystyle \sum _{i=1}^{n}fi}=50$ .

So, mean,  $\overline{x}=\frac{1}{N}*{\displaystyle \sum _{i=1}^{n}fi}xi=\frac{1350}{5}=27.$

$\begin{array}{c}=\frac{1}{50}*6600\\ \end{array}$

= 132.

19. Let the assumed mean be A=105 and class width, h=30. The given data can be tabulated as

= 2280 - 4

= 2276.

18. Let the assumed mean be A=64 and it the width, h=1.