Mathematics, 27.11.2019 02:31 shaakiraak

Almost all medical schools in the united states require students to take the medical college admission test (mcat). to estimat the mean score μμ of those who took the mcat at etown, you obtain a simple random sample of 26 students' scores. the scores follow a normal distribution, and from published information you know that the standard deviation is 6.4. suppose that (unknown to you) the mean score of those taking the mcat at etown is 25.

(a) if you choose one mcat score at random, then what is the probability that the student's score is between 20 and 30?

(b) what is the probability that the mean score of your sample is between 20 and 30?

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Answer from: laciekersey4568

a) There is a 56.46% probability that the student's score is between 20 and 30.

b) There is a 100% probability that the mean score of your sample is between 20 and 30.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Problems of normally distributed samples can be 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:

The scores follow a normal distribution, and from published information you know that the standard deviation is 6.4. Suppose that (unknown to you) the mean score of those taking the MCAT at Etown is 25. This means that $\mu = 25, \sigma = 6.4$ .

(a) If you choose one MCAT score at random, then what is the probability that the student's score is between 20 and 30?

This probability is the pvalue of Z when X = 20 subtracted by the pvalue of Z when X = 30 . So

X = 30

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

$Z = \frac{30 - 25}{6.4}$

Z = 0.78

Z = 0.78 has a pvalue of 0.7823

X = 20

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

$Z = \frac{20 - 25}{6.4}$

Z = -0.78

Z = -0.78 has a pvalue of 0.2177

This means that there is a 0.7823 - 0.2177 = 0.5646 = 56.46% probability that the student's score is between 20 and 30.

(b) What is the probability that the mean score of your sample is between 20 and 30?

By the Central Limit Theorem, now we have x = 26 and $s = \frac{6.4}{\sqrt{26}}$

X = 30

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

$Z = \frac{30 - 25}{1.255}$

Z = 3.98

Z = 3.98 has a pvalue of 1

X = 20

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

$Z = \frac{20 - 25}{6.4}$

Z = -3.98

Z = -3.98 has a pvalue of 0

This means that there is a 100% probability that the mean score of your sample is between 20 and 30

Answer from: Quest

the solution is the point $(1,-3)$

step-by-step explanation:

we have

$3x-y=6$ > equation a

$y=x-4$ > equation b

adds equation a and equation b

$3x-y=6\\y=x--y+y=6+x-4\\3x=2+x\\3x-x=2\\2x=2\\x=1$

substitute the value of x in equation b to find the value of y

$y=(1)-4=-3$

the solution is the point $(1,-3)$

Answer from: Quest

2/3

step-by-step explanation:

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