Question

for a sample of 117 transformers built for heavy industry, the mean and standard deviation of the number of sags per week were 370 and 26, respectively. also, the mean and standard deviation of the number of swells per week were 174 and 23, respectively. consider a transformer that has 277 sags and 208 swells in a week. complete parts a and b below. b. would you consider 208 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 , meaning that less than approximately 68% of transformers have a number of swells closer to the mean. (round to two decimal places as needed.) b. yes. the z - score is , meaning that this is an outlier and almost every other transformer has more swells. (round to two decimal places as needed.) c. yes. the z - score is , meaning that this is an outlier and almost every other transformer has fewer swells. (round to two decimal places as needed.) d. no. the z - score is , meaning that the number of swells is not unusual and is not an outlier. (round to two decimal places as needed.)

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.
We are given $\mu = 174$ (mean number of swells per week), $\sigma=23$ (standard deviation of the number of swells per week), and $x = 208$ (the number of swells we are interested in).

Step2: Calculate the z - score

Substitute the values into the formula: $z=\frac{208 - 174}{23}=\frac{34}{23}\approx1.48$.

Step3: Determine if it's unusual

In general, a z - score with $|z|> 2$ is considered unusual. Since $|1.48|<2$, the number of swells is not unusual.

Answer:

A. No. The z - score is $1.48$, meaning that less than approximately 68% of transformers have a number of swells closer to the mean.