Question

The weights of 1500 ball bearings are normally distributed with a mean of 635 gms and S.D. of 1.36 gms. If 300 random samples of size 36 are drawn from this population. In the case of random sampling with replacement, find how many random samples would
have their mean
(i) between 634.76gms and 635.24 gm
(ii)
less than 635.6 gms
(iii) greater than 635.6
gms,
(iv) less than 634.5 gms or more than 635.24 gms
[P(1.06)=0.3554, P(2.65)=0.4960, P(2.21)=0.4864, P(1.06)=0.3554]

Answer 1

Answer:

Between 634,76 gms and 635,24

Answer 2

Given :  The weights of 1500 ball bearings are normally distributed with a mean of 635 gms and S.D. of 1.36 gms. If 300 random samples of size 36 are drawn from this population. In the case of random sampling with replacement,

To find : how many random samples would  have their mean

Solution:

a mean of 635 gms and S.D. of 1.36 gms.  

samples of size 36

SE = SD/√n    =   1.36 / √36  =  0.2267

between 634.76gms and 635.24 gm

z = (634.76 - 635)/ 0.2267     and  (635.24 - 635)/ 0.2267

=>  z = -1.06   and   1.06

between 634.76gms and 635.24 gm = 0.8554  -  0.1446  = 0.7108

0.7108 of 300  =   213.24   ≈ 213

greater than 635.6 gms,  

z =  (635.6 - 635)/ 0.2267  = 2.65

 1- 0.9960 =  0.0040

0.0040 of 300  = 1.2 ≈ 1

less than 635.6 gms = 0.9960 of 300 = 298.8 ≈ 299

less than 634.5 gms   or more than 635.24 gms

more than 635.24 gms  = 1 - 0.8554  = 0.1446  of 300 =  43.38 ≈  43

less than 634.5  

z =   (634.5 - 635)/ 0.2267  = -2.21

=>  0.0136  of 300 = 4.08   ≈  4

Hence 43 + 4 = 7    less than 634.5 gms or more than 635.24 gms

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