For a sample of transformers built for heavy industry, the mean and standard deviation of the number of sags per weeks were and , respectively; also, the mean and standard deviation of the number of swells per week were and , respectively. Consider a transformer that has sags and swells in a week. Complete parts a and b below.
1. Would you consider 314 sags per week unusual, statistically? Explain.
A. No. The Z-score is meaning that the number of sags is not unusual and is not an outlier.
B. No. The Z-score is meaning that less than approximately 68% of transformers have a number of sags closer to the mean.
C. Yes. The Z-score is meaning that this is an outlier and almost every other transformer has fewer sags.
D. Yes. The Z-score is meaning that this is an outlier and almost every other transformer has more sags.
E. This cannot be determined, since the IQR is not provided and cannot be found from the provided information.
2. Would you consider 243 swells per week unusual, statistically? Explain.
A. No. The Z-score is meaning that less than approximately 68% of transformers have a number of swells closer to the mean.
B. Yes. The Z-score is meaning that this is an outlier and almost every other transformer has more swells.
C. Yes. The z-score is meaning that this is an outlier and almost every other transformer has fewer swells.
D. No. The Z-score is meaning that the number of swells is not unusual and is not an outlier.
E. This cannot be determined, since the IQR is not provided and cannot be found from the provided information.

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 z-score less than -2 or more than 2 are considered as unusual.

Compute the z-score for 314 sags per week as follows:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

[tex]=\frac{314-338}{22}\\\\=-1.091[/tex]

The z-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 -1.091 meaning that the number of sags is not unusual and is not an outlier.

(2)

Compute the z-score for 243 swells per week as follows:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

[tex]=\frac{243-174}{15}\\\\=4.6[/tex]

The z-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 4.6 meaning that this is an outlier and almost every other transformer has fewer swells.

JackelineCasarez JackelineCasarez · Answer 2 · 2022-01-20T15:57:47+01:00

1). The decision regarding the consideration of 314 sags every week unusually would be:

a). No. The Z-score is meaning that the number of sags is not unusual and is not an outlier.

2). The decision regarding the consideration of 243 swells every week every week unusually would be:

c). Yes. The z-score is meaning that this is an outlier and almost every other transformer has fewer swells.

a). Given that

No. of transformers = 119

Mean = 338

Standard Deviation for the week = 22

Now,

z-score for sags = (No. of sags - Mean)/Standard Deviation

[tex]= (314 - 338)/22\\= -1.091[/tex]

Since it is greater than -2, option A is the correct answer.

b). Given that

No. of swells = 243

Mean for swells = 174

Standard Deviation = 15

Now,

z-score for sags = (No. of sags - Mean)/Standard Deviation

[tex]= (243 - 174)/15\\= 4.6[/tex]

Since it is greater than 2, it exemplifies an outlier. Thus, option C is the correct answer.

Thus, options A and C are the correct answers.

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