Question

for a sample of 120 transformers built for heavy industry, the mean and standard deviation of the number of sags per week were 327 and 27, respectively. also, the mean and standard deviation of the number of swells per week were 178 and 18, respectively. consider a transformer that has 362 sags and 116 swells in a week. complete parts a and b below. a. would you consider 116 swells per week unusual, statistically? explain. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. no. the z - score is (round to two decimal places as needed.), meaning that less than approximately 68% of transformers have a number of swells closer to the mean. b. no. the z - score is (round to two decimal places as needed.), meaning that the number of swells is not unusual and is not an outlier. c. yes. the z - score is (round to two decimal places as needed.), meaning that this is an outlier and almost every other transformer has fewer swells. d. yes. the z - score is (round to two decimal places as needed.), meaning that this is an outlier and almost every other transformer has more swells.

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x - \mu}{\sigma}$, where $x$ is the data point, $\mu$ is the mean, and $\sigma$ is the standard deviation. For the number of swells, $\mu = 18$, $\sigma=178$, and $x = 116$.

Step2: Calculate the z - score

$z=\frac{116 - 18}{178}=\frac{98}{178}\approx0.55$.

Step3: Determine if it's unusual

In a normal distribution, values with $|z|> 2$ or $|z|>3$ are considered unusual. Since $|0.55|<2$, the number of swells is not unusual and not an outlier.

Answer:

B. No. The z - score is $0.55$, meaning that the number of swells is not unusual and is not an outlier.