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
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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