Answer:
The correct options are:
(1) A
(2) C
Step-by-step explanation:
The complete question is:
For a sample of 119 transformers built for heavy industry, the mean and standard deviation of the number of sags per weeks were 338 and 22, respectively; also, the mean and standard deviation of the number of swells per week were 174 and 15, respectively. Consider a transformer that has sags 375 and 120 swells in a week.
(1)
A <em>z</em>-score less than -2 or more than 2 are considered as unusual.
Compute the <em>z</em>-score for 314 sags per week as follows:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
![=\frac{314-338}{22}\\\\=-1.091](https://tex.z-dn.net/?f=%3D%5Cfrac%7B314-338%7D%7B22%7D%5C%5C%5C%5C%3D-1.091)
The <em>z</em>-score for 314 sags per week is -1.091.
This value is more than -2.
Thus, the correct option is:
A. No. The Z-score is <u>-1.091</u> meaning that the number of sags is not unusual and is not an outlier.
(2)
Compute the <em>z</em>-score for 243 swells per week as follows:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
![=\frac{243-174}{15}\\\\=4.6](https://tex.z-dn.net/?f=%3D%5Cfrac%7B243-174%7D%7B15%7D%5C%5C%5C%5C%3D4.6)
The <em>z</em>-score for 243 swells per week is 4.6.
This value is more than 2.
Thus, the correct option is:
C. Yes. The z-score is <u>4.6</u> meaning that this is an outlier and almost every other transformer has fewer swells.