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1 answer:
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Answer:
(a) 0.119
(b) 0.1699
Solution:
As per the question:
Mean of the emission, million ponds/day
Standard deviation, million ponds/day
Now,
(a) The probability for the water pollution to be at least 15 million pounds/day:
= 1 - P(Z < 1.178)
Using the Z score table:
= 1 - 0.881 = 0.119
The required probability is 0.119
(b) The probability when the water pollution is in between 6.2 and 9.3 million pounds/day:
P(Z < - 0.86) - P(Z < - 1.96)
Now, using teh Z score table:
0.1949 - 0.025 = 0.1699