Question

According to an article in Newsweek, the rate of water pollution in China is more than twice that measured in the US and more than three times the amount measured in Japan. The mean emission of organic pollutants is 11.7 million pounds per day in China. Assume the water pollution in China is normally distributed throughout the year with a standard deviation of 2.8 million pounds of organic emissions per day. a) What is the probability that on any given day the water pollution in China is at least 15 million pounds per day? b) What is the probability that on any given day the water pollution in China is between 6.2 and 9.3 million pounds per day?

Answer 1

Answer:

(a) 0.119

(b) 0.1699

Solution:

As per the question:

Mean of the emission, $\mu = 11.7$ million ponds/day

Standard deviation, $\sigma = 2.8$ million ponds/day

Now,

(a) The probability for the water pollution to be at least 15 million pounds/day:

$P(X\geq 15) = P(\frac{X - /mu}{\sigma} \geq \frac{15 - 11.7}{2.8})$

$P(X\geq 15) = P(Z \geq 1.178)$

$P(X\geq 15)$ = 1 - P(Z < 1.178)

Using the Z score table:

$P(X\geq 15)$ = 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:

$P(6.2 < X < 9.3) = P(\frac{6.2 - 11.7}{2.8} < \frac{X - \mu}{\sigma} < \frac{9.3 - 11.7}{2.8})$

$P(6.2 < X < 9.3) = P(- 1.96 < Z < - 0.86)$

$P(6.2 < X < 9.3) = P(- 1.96 < Z < - 0.86)$

P(Z < - 0.86) - P(Z < - 1.96)

Now, using teh Z score table:

0.1949 - 0.025 = 0.1699