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Payal Gupta

Contributor-Level 10

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

We have mean = x¯=i=1nxin(1)

and variance = σ2=1ni=1n(xix¯)2(2)

Let yi  be the new observations with same n.

So,  yi = axi      (3)

Now mean, y¯=i=1nyin=i=1naxin=ai=1nyin=ax¯[(1)]

So  y¯=ax¯(4)

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

σ2=1ni=1n(yiay¯a)2

σ2=1a2[1ni=1n(yiy¯)2]

(σ')2=σ2α2.

Hence, the mean and variance of ax1, ax2, ……, axn  are ax¯ and a2 σ2 .

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Payal Gupta

Contributor-Level 10

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

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Payal Gupta

Contributor-Level 10

27. Given, n=50.

i=1nxi=212i=150xi2=902.8

i=150yi=261i=150yi2=1457.6

So,  x¯=i=150xin=21250=4.24

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Payal Gupta

Contributor-Level 10

26. For team A.

Hence, we conclude that team A is more consistent.

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Payal Gupta

Contributor-Level 10

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

Total no of were canner in firm B.

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

=? 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.

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Payal Gupta

Contributor-Level 10

24. We can tabulate the given data as follows.

Total =

Mean of x, x¯=xin=51010=51.

Mean of Y, Y¯=Yin=105010=105 .

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Payal Gupta

Contributor-Level 10

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

=6150*10+45.

= 44.6

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