Using z-scores, a population with µ = 37 and σ = 6 is standardized so that the new mean is µ = 50 and σ = 10. After the standardization, one individual has a score of X = 55. What was this individual’s score in the original distribution?

X = 43
It cannot be determined with the information given.
X = 40
X = 42

Respuesta :

Answer:

X = 40

Step-by-step explanation:

Original mean (µ) = 37

Original standard deviation (σ) = 6

New mean (µ) = 50

New standard deviation (σ) = 10

The z-score of an individual who had a score of X=55 in the new distribution is given by:

[tex]z=\frac{X-\mu}{\sigma}\\z=\frac{55-50}{10}\\z=0.5[/tex]

A z-score of 0.5, in the original distribution yields a score, X, of:

[tex]0.5=\frac{X-37}{6}\\X=0.5*6 +37\\X=40[/tex]

This individual’s score in the original distribution is X = 40

Following are the calculation for the distribution:

Given:

Original value

[tex]\mu=37\\\\\sigma=6[/tex]

New value

[tex]\mu=50\\\\\sigma=10[/tex]

[tex]X=55[/tex]

To find:

original distribution [tex](X)=?[/tex]

Solution:

In the given question, when the z-score is an individual then the [tex]X=55[/tex] so, the new distribution:

[tex]\to Z=\frac{X-\mu}{\sigma}=\frac{55-50}{10}=\frac{5}{10}=0.5[/tex]

Now we using the Z-score value that is 0.5. so, the original distribution of the X score:

[tex]\to 0.5=\frac{X-37}{6}\\\\\to X=0.5 \times 6 +37\\\\\to X=3.0+37\\\\\to X= 40[/tex]

So, the individual’s score for the original distribution of "[tex]X = 40[/tex]".

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