Maths

2. – 12 x> 30.

Dividing both sides by 12 we get,

$-x>\frac{30}{12}$

$-x>\frac{5}{2}$

Multiplying both side by ( – 1) the inequality will change.

i e, x< $\frac{5}{2}$ = – 2.5

(i) As x is a natural number, the soln of the given inequality does not exist in natural numbers.

(ii) As x is an integer, the soln of the given inequality will be all the integer less than – 5/2

i. e, – 3, – 4, – 5, ….

1. 24x < 100.

Dividing both sides by 24 we get,

$\frac{24x}{24}<\frac{150}{24.}$

x< $\frac{25}{6}$ = 4.166

(i) As x is a natural no the solⁿ of the given inequality are 1, 2, 3, 4.

(ii) As x is an integer the solⁿ of the given inequality are 4, 3, 2, 1, 0, 1, 2, 3, ….

34. Given, n = 100.

incorrect mean ( $\overline{x}$ ) = 20.

incorrect standard deviation (σ) = 3

We know that,

$\overline{x}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{n}_i}$

$\Rightarrow 20=\frac{1}{100}{\displaystyle \sum _{i=1}^n{n}_i}$

$\Rightarrow {\displaystyle \sum _{i=1}^n{n}_i}=2000.$

So, incorrect sum of observation = 2000

Correct sum of observation $=2000-21-21-18$

= 1940

33. C.V in mathematics = $\frac{12}{42}*100=28.57.$

C.V in Physics = $\frac{15}{32}*100=46.87.$

C.V in chemistry = $\frac{20}{40.9}*100=48.89$

$\therefore$  Chemistry has the highest variability and mathematics has the lowest variability.

32.  (i) Given, n = 20.

Incorrect mean $(\overline{x})=10$

Incorrect standard deviation $(\sigma )=2$

We know that,

$\overline{x}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}$

$\Rightarrow 10=\frac{1}{20}{\displaystyle \sum _{i=1}^{20}{x}_i}$

$\Rightarrow {\displaystyle \sum _{i=1}^{20}{x}_i}=200$

So, incorrect sum of observation = 200.

correct sum of observation =200 – 8 = 192

And correct mean $=\frac{192}{19}=10.1$

31. For n observations x₁, x₂,……., x_n .

We have mean = $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\left(1\right)$

and variance = ${\sigma }^2=\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2\left(2\right)}$

Let y_i  be the new observations with same n.

So,  y_i = ax_i      (3)

Now mean, $\overline{y}=\frac{{\displaystyle \sum _{i=1}^n{y}_i}}{n}=\frac{{\displaystyle \sum _{i=1}^{n}a{x}_i}}{n}=\frac{a{\displaystyle \sum _{i=1}^n{y}_i}}{n}=a\overline{x}\left[\left(1\right)\right]$

So  $\overline{y}=a\overline{x}\left(4\right)$

And, putting (3) and (4) in (2) we get,

${\sigma }^2=\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(\frac{y_i}{a}-\frac{\overline{y}}{a}\right)}^2}$

$\Rightarrow {\sigma }^2=\frac{1}{a^2}\left[\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}\right]$

$\Rightarrow {\left({\sigma }^{\text{'}}\right)}^2={\sigma }^2{\alpha }^2.$

Hence, the mean and variance of ax₁, ax₂, ……, ax_n  are $a\overline{x}$ and a² σ² .

30. Let 'x' be the given observations with n = 6.

27. Given, n=50.

$\begin{array}{l}\\ {\displaystyle \sum _{i=1}^{n}xi}=212{\displaystyle \sum _{i=1}^{50}x{i}^2}=902.8\\ \end{array}$

${\displaystyle \sum _{i=1}^{50}yi}=261{\displaystyle \sum _{i=1}^{50}y{i}^2}=1457.6$

So,  $\overline{x}=\frac{{\displaystyle \sum _{i=1}^{50}{x}_i}}{n}=\frac{212}{50}=4.24$

26. For team A.

Hence, we conclude that team A is more consistent.

(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.