Question

Employees that work at a fish store must measure the level of nitrites in the water each day. Nitrite levels should remain lower than 5 ppm as to not harm the fish. The nitrite level varies according to a distribution that is approximately normal with a mean of 3 ppm.

The probability that the nitrite level is less than 2 ppm is 0.0918. Which of the following is closest to the probability that on a randomly selected day the nitrite level will be at least 5 ppm?

A .0039
B .0266
C .0918
D .7519
E .9961

These are subjects that I haven't done in a long time so if you could provide work too, that would be amazing!

Answer 1

0.0039

Explanation:

We must first determine the standard deviation of the distribution. The z-score corresponding to a left tail area of 0.0918 is z = –1.33. We must solve –1.33 = (2 – 3). Therefore = 0.7519. We need to find P(X ≥ 5). The z-score for X = 5 is z = (5 – 3)/0.7519 = 2.66. Using Table A, the area to the left of z = 2.66 is 0.9961, so the area to the right of z = 2.66 is 1 – 0.9961 = 0.0039. P(X ≥ 5) = 0.0039.

Answer 2

Answer:.0039

Step-by-step explanation: we must first determine the sd of the distribution. The z score corresponding to a left tail area of .0918 is z=-1.33. We must solve -1.33=(2-3)/sd for the sd, which is .7519. Then use that to the probability of p(x>5).