Math, asked by magishakumar07, 3 months ago

The life lengths in hours of 2 electronic devices A and B have distributions N (40, 6) and N (45, 3) respectively. If the electronic device is to be used for a 45 –h period, which device is to be preferred? If it is to be used for a 48-h period, which device is to be preferred?

Answers

Answered by amitnrw
0

Given : The life lengths in hours of 2 electronic devices A and B have distributions N (40, 6) and N (45, 3) respectively.

To Find : If the electronic device is to be used for a 45 –h period, which device is to be preferred?

If it is to be used for a 48-h period, which device is to be preferred?

Solution:

Device A  

N ( 40 , 6)

Mean = 40

SD = 6

device is to be used for a 45 –h period

Z score = (45 - 40)/6 = 5/6    = 0.833  => 0.7975  

Hence 1 - 0.7975  = 0.2025 above 45 hrs

device is to be used for a 48 –h period

Z score = (48 - 40)/6 = 8/6    = 1.333  => 0.9085  

Hence 1 -  0.9085   = 0.0915 above 48 hrs

Device B

N ( 45 , 3)

Mean = 45

SD = 3

device is to be used for a 45 –h period

Z score = (45 - 45)/3 = 0    = 0.5

Hence 1 - 0.5  = 0.5 above 45 hrs

0.5 > 0.2025  Hence  Device B is better for 45 hours

device is to be used for a 48 –h period

Z score = (48 - 45)/3 = 3/3    = 1  => 0.8413

Hence 1 - 0.8413   = 0.1587 above 48 hrs

0.1587  >  0.0915

Hence Device B is better for 48 hours

in Both cases Device B is better

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