4.2x−1.4y=2.1 in terms of y

1 answer:

3 0

To find Y we must cancel X out by making it 0.

4.2(0) - 1.4y = 2.1

-1.4y = 2.1

Divide both sides by -1.4 to get the Y by itself.

Y = 2.1/-1.4
Y = -1.5

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The total scores on the Medical College Admission Test (MCAT) in 2013 follow a Normal distribution with mean 25.3 and standard d

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In a normal distribution, the median is the same as the mean (25.3). The first quartile is the value of $Q_1$ such that

$\mathbb P(X$

You have

$\mathbb P(X$

For the standard normal distribution, the first quartile is about $z\approx-0.6745$ , and by symmetry the third quartile would be $z\approx0.6745$ . In terms of the MCAT score distribution, these values are

$\dfrac{Q_1-25.3}{6.5}=-0.6745\implies Q_1\approx20.9$
$\dfrac{Q_3-25.3}{6.5}=0.6745\implies Q_3\approx29.7$

The interquartile range (IQR) is just the difference between the two quartiles, so the IQR is about 8.8.

The central 80% of the scores have z-scores $\pm z$ such that

$\mathbb P(-z$

That leaves 10% on either side of this range, which means

$\underbrace{\mathbb P(-z$

You have

$\mathbb P(Z$

Converting to MCAT scores,

$-1.2816=\dfrac{x_{\text{low}}-25.3}{6.5}\implies x_{\text{low}}\approx17.0$
$1.2816=\dfrac{x_{\text{high}}-25.3}{6.5}\implies x_{\text{high}}\approx33.6$

So the interval that contains the central 80% is $(17.0,33.6)$ (give or take).