Question

Many urban zoos are looking at ways to effectively handle animal waste. One zoo has installed a facility that will transform animal waste into electricity. To estimate how many pounds of waste they may have to fuel the new facility they began keeping meticulous records. They discovered that the amount of animal waste they were disposing of daily is approximately Normal with a mean of 348.5 pounds and a standard deviation of 38.2 pounds. Amounts over 350 pounds would generate enough electricity to cover what is needed to for the entire aquarium that day. Approximately what proportion of the days can the zoo expect to obtain enough waste to cover what is needed to run the entire aquarium for the day (A) 0.484 (B) 0.499 (C) 0.516 (D) 0.680 (E) 0.950

Answer 1

Answer:

• Option A: 0.484

Explanation:

The amount of animal waste one zoo is diposing daily is approximately normal with:

• mean, μ = 348.5 lbs
• standard deivation, σ = 38.2 lbs

The proportion of waste over 350 lbs may be found using the table for the area under the curve for the cumulative normal standard probability.

First, find the z-score for 350 lbs:

      $z-score=\dfrac{X-\mu}{\sigma}$

      $z-score=\dfrac{350lbs-348.5lbs}{38.2lbs}\approx0.04$

There are tables for the cumulative areas (probabilities) to the left and for the cumulative areas to the right of the z-score.

You want the proportion of the days when the z-score is more than 0.04; then, you can use the table for the values to the rigth of z = 0.04.

From such table, the area or probability is 0.4840.

The attached image shows a portion of the table with that value: it is the cell highlighted in yellow.

Hence, the answer is the option (A) 0.484.