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) 0.0039

(B) 0.0266

(C) 0.0918

(D) 0.7519

(E) 0.9961

Answer 1

Answer:

$-1.330 = \frac{2-3}{\sigma}$

And olving for the deviation we got:

$\sigma = \frac{2-3}{-1.330} = 0.752$

And then we want to find this probability:

$P(X>5)$

Using the z score we got:

$P(X>5)= P(Z>\frac{5-3}{0.752}) = P(Z>2.660)$

And using the complement rule and the normal standard distirbution or excel we got:

$P(X>5)= P(Z>2.660)= 1-P(Z<2.660) = 1-0.99609 =0.00391$

And the best answer for this case is:

(A) 0.0039

Step-by-step explanation:

For this case we define the random variable X as " The nitrite level" and we know that the distribution of X is given by:

$X \sim N (\mu =3, \sigma )$

We don't know the deviation. We also know the following condition:

$P(X<2) = 0.0918$

So then we can use the z score formula given by:

$z= \frac{X -\mu}{\sigma}$

We need to find a z score value that accumulates 0.0918 of the area on the right and we got z = -1.330, since P(Z<-1.330) = 0.0918. Then we can set up the following equation:

$-1.330 = \frac{2-3}{\sigma}$

And olving for the deviation we got:

$\sigma = \frac{2-3}{-1.330} = 0.752$

And then we want to find this probability:

$P(X>5)$

Using the z score we got:

$P(X>5)= P(Z>\frac{5-3}{0.752}) = P(Z>2.660)$

And using the complement rule and the normal standard distirbution or excel we got:

$P(X>5)= P(Z>2.660)= 1-P(Z<2.660) = 1-0.99609 =0.00391$

And the best answer for this case is:

(A) 0.0039