Question

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. Refer to Exhibit 4. What percentage of items will weigh at the most 11.7 ounces?

a) 84%
b) 93%
c) 97.5%
d) 99.7%

Answer 1

To find the answer, calculate the z-score for 11.7 ounces. Then reference a standard normal distribution table or a calculator to determine the percentage which is approximately 96.8%, with the closest provided option being 93%.

Step-by-step explanation:

To find the percentage of items from a normally distributed set that weigh at most 11.7 ounces, with a mean of 8 ounces and a standard deviation of 2 ounces, we need to calculate the z-score and then refer to a standard normal distribution table or use a calculator with normal distribution capabilities.

First, we calculate the z-score using the formula: z = (X - μ) / σ where X is the value in question, μ is the mean, and σ is the standard deviation. Plugging in the numbers:

z = (11.7 - 8) / 2 = 3.7 / 2 = 1.85

Next, we look up the cumulative probability for z = 1.85 in the z-table or use a calculator to find that it corresponds to approximately 96.8%. Since the options provided do not include this exact percentage, we select the option that is closest without going over. Therefore, the correct answer is b) 93%.

Remember, the percentage from a z-table or a calculator provides the probability that a value is less than or equal to our given value of 11.7 ounces in a normal distribution.