In a certain state, the daily amounts of industrial waste are normally distributed, with a mean of 6,000 tons and a standard deviation of 2,000 ons. Find the probability that on a randomly selected day the amount of industrial waste is between 9,400 and 11,800 tons. Use a table of the standard normal distribution to solve the problem. Round your answer to four decimal places.
A)0.0427
B)0.9573
C)0.2911
D)0.4270
E)0.4697

Question

In a certain state, the daily amounts of industrial waste are normally distributed, with a mean of 6,000 tons and a standard deviation of 2,000 ons. Find the probability that on a randomly selected day the amount of industrial waste is between 9,400 and 11,800 tons. Use a table of the standard normal distribution to solve the problem. Round your answer to four decimal places. ​
A)0.0427
B)0.9573
C)0.2911
D)0.4270
E)0.4697

Quizplus · Accepted Answer

To find the probability that the amount of industrial waste is between 9,400 and 11,800 tons, we first convert these values to Z-scores using the formula $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. For 9,400 tons, $Z = \frac{9400 - 6000}{2000} = 1.7$. For 11,800 tons, $Z = \frac{11800 - 6000}{2000} = 2.9$. Using a standard normal distribution table, we find the probabilities corresponding to these Z-scores and subtract the smaller from the larger to find the probability of being between these two values. The probability corresponding to Z=1.7 is approximately 0.9554, and for Z=2.9, it is approximately 0.9981. The difference, 0.9981 - 0.9554, gives approximately 0.0427, which matches option A.

In a Certain State, the Daily Amounts of Industrial Waste