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Find the z-score corresponding to the given value and use the z-score to determine whether the value is unusual. Consider a score to be unusual if its z-score is less than -2.00 or greater than 2.00. Round the z-score to the nearest tenth if necessary. A weight of 220 pounds among a population having a mean weight of 161 pounds and a standard deviation of 23.5 pounds Group of answer choices

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Answer:

The z-score is 2.5106 which is an unusual value of z-score as it is greater than 2.00

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 161 pounds

Standard Deviation, σ = 23.5 pounds

x = 220 pounds

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

Putting values, we get,

[tex]z_{score}=\dfrac{220-161}{23.5}=2.5106[/tex]

Thus, the z-score is 2.5106 which is an unusual value of z-score as it is greater than 2.00

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