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
(8) 0.0266
(C) 0.0918
(D) 0.7519
(5) 0.9961

Answer 1

Answer:

(A) 0.0039

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3$

The probability that the nitrite level is less than 2 ppm is 0.0918.

This means that when $X = 2$, Z has a pvalue of 0.0918. So when X = 2, Z = -1.33.

We use this to find $\sigma$

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

$-1.33 = \frac{2 - 3}{\sigma}$

$-1.33\sigma = -1$

$1.33\sigma = 1$

$\sigma = \frac{1}{1.33}$

$\sigma = 0.7519$

Which of the following is closest to the probability that on a randomly selected day the nitrite level will be at least 5 ppm

This is 1 subtracted by the pvalue of Z when X = 5. So

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

$Z = \frac{5 - 3}{0.7519}$

$Z = 2.66$

$Z = 2.66$ has a pvalue of 0.9961

1 - 0.9961 = 0.0039

So the correct answer is:

(A) 0.0039

Answer 2

Answer:

(A) 0.0039

Step-by-step 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)/\sigmaσ, for \(\sigma\). 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.