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There are two major tests of readiness for college: the act and the sat. act scores are reported on a scale from 1 to 36. the distribution of act scores in recent years has been roughly normal with mean μ = 20.9 and standard deviation σ = 4.8. sat scores are reported on a scale from 400 to 1600. sat scores have been roughly normal with mean μ = 1026 and standard deviation σ = 209. joe scores 1351 on the sat. assuming that both tests measure the same thing, what score (±1) on the act is equivalent to joe's sat score? .

Respuesta :

Solution: We are given:

ACT scores follow normal distribution with [tex]\mu=20.9,\sigma =4.8[/tex]

SAT scores follow normal distribution with [tex]\mu=1026,\sigma=209[/tex]

Now, let's find the z score corresponding to Joe's SAT score 1351.

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

        [tex]=\frac{1351-1026}{209}[/tex]

        [tex]=\frac{325}{209}[/tex]

        [tex]=1.56[/tex]

Therefore, Joe's SAT score is 1.56 standard deviations above the mean.

Now, we have find the Joe's ACT score, which will be 1.56 standard deviations above the mean.

Therefore, we have:

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

[tex]1.56=\frac{x-20.9}{4.8}[/tex]

[tex]1.56 \times 4.8 = x - 20.9[/tex]

[tex]7.488=x-20.9[/tex]

[tex]x=20.9+7.488[/tex]

[tex]x=28.388 \approx 28.4[/tex]

Therefore, Joe's equivalent ACT score to SAT score 1351 is 28.4    

You can use the standard normal variate as an intermediary for both the distributions' score translation for same person.

The test score of Joe in act test which is equivalent to his sat score, is 28.39 approximately.

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

[tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean[tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

Let X be the random variable tracking the act scores and Y be the random variable tracking the sat scores of students of specified college.

Then, by the given data, we have:

[tex]X \sim N(20.9, 4.8)\\\\Y \sim N(1026, 209)[/tex]

Since it is given that both the score measure the same thing, that means both the scores are just varying in scale and origin shift from each other. So one solution might be to standardize them. Using standard normal distribution, we have:

[tex]Z = \dfrac{X - 20.9}{4.8}\\\\Z = \dfrac{Y - 1026}{209}\\[/tex]

Since Joe scored 1351 marks in sat exam, thus, its Z score will be

[tex]Z = \dfrac{Y - 1026}{209} = \dfrac{1351- 1026}{209} \approx 1.56[/tex]

Both scores measure same thing, so on standard distribution, there scores should be same, thus,

[tex]Z = \dfrac{X - 20.9}{4.8} = \dfrac{Y - 1026}{209}\\\\\\1.56 = \dfrac{X - 20.9}{4.8} \\\\X = 1.56 \times 4.8 + 20.9 \approx 28.39[/tex]

Thus,

The test score of Joe in act test which is equivalent to his sat score, is 28.39 approximately.

Learn more about standard normal distribution here:

https://brainly.com/question/21262765

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