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3 years ago

X –27 = 24 so X = ?

Mathematics

1 answer:

Alexeev081 [22]3 years ago

6 0

Answer:

X = 51

Step-by-step explanation:

24 + 27 = 51

x - 27 =

51 - 27 = 24

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A college requires applicants to have an ACT score in the top 12% of all test scores. The ACT scores are normally distributed, w

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Answer:

a) The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b) 156 would be expected to have a test score that would meet the colleges requirement

c) The lowest score that would meet the colleges requirement would be decreased to 26.388.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 21.5, \sigma = 4.7$

a. Find the lowest test score that a student could get and still meet the colleges requirement.

This is the value of X when Z has a pvalue of 1 - 0.12 = 0.88. So it is X when Z = 1.175.

$Z = \frac{X - \mu}{\sigma}$

$1.175 = \frac{X - 21.5}{4.7}$

$X - 21.5 = 1.175*4.7$

$X = 27.0225$

The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b. If 1300 students are randomly selected, how many would be expected to have a test score that would meet the colleges requirement?

Top 12%, so 12% of them.

0.12*1300 = 156

156 would be expected to have a test score that would meet the colleges requirement

c. How does the answer to part (a) change if the college decided to accept the top 15% of all test scores?

It would decrease to the value of X when Z has a pvalue of 1-0.15 = 0.85. So X when Z = 1.04.

$Z = \frac{X - \mu}{\sigma}$

$1.04 = \frac{X - 21.5}{4.7}$

$X - 21.5 = 1.04*4.7$

$X = 26.388$

The lowest score that would meet the colleges requirement would be decreased to 26.388.

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