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jasenka [17]
3 years ago
15

A new version of the Medical College Admissions Test (MCAT) was introduced in spring 2015 and is intended to shift the focus fro

m what applicants know to how well they can use what they know. One result of the change is that the scale on which the exam is graded has been modified, with the total score of the four sections on the test ranging from 472 to 528. In spring 2015, the mean score was 500.0 with a standard deviation of 10.6.
What are the median and the first and third quartiles of the MCAT scores?

Median=
Q1(rounded to three decimal places)=
Q3(rounded to three decimal places)=
The IQR (interquartile range, rounded to three decimal places) =

Choose the interval that contains the central 80% of the MCAT scores.

a.491.096 to 508.904
b.472 to 528
c.480 to 520
d.486.432 to 513.568
Mathematics
1 answer:
Jet001 [13]3 years ago
3 0

Answer:

  • median=500
  • Q1= 492.792
  • Q3= 507.208
  • IQR= 14.416

central 80% of the MCAT scores is 486.432 to 513.568

Step-by-step explanation:

If we assume a normal distribution of the MCAT scores in spring 2015, then mean=median=500

first quartile Q1 is the first 25% boundary of the scores, thus

P(z<z*)=0.25 where z* is the z-score of Q1.

P(z<z*)=0.25  gives z*=-0.68

z-score of Q1 can be stated as follows:

-0.68=\frac{X-M}{s} where where

  • X  is the Q1 score
  • M is the mean  MCAT score (500)  
  • s is the standard deviation (10.6)

Solving the equation for Q1 we get Q1= -0.68×10.6 + 500 ≈ 492.792

Similarly for third quartile Q3  z* is  P(z<z*)=0.75 gives z*=0.68

That is Q3= 0.68×10.6 + 500 ≈ 507.208

IQR ( Interquartile Range) is IQR=Q3-Q1 = 507.208 - 492.792 = 14.416

for central 80% of the MCAT scores P(-z*<z<z*)=0.80 gives z*=1.28

that is upper bound = 1.28×10.6 + 500 ≈ 513.568

lower bound = -1.28×10.6 + 500 ≈ 486.432

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Decide if the following statement is valid or invalid. If two sides of a triangle are congruent then the triangle is isosceles.
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Answer:

Step-by-step explanation:

Properties of an Isosceles Triangle

(Most of this can be found in Chapter 1 of B&B.)

Definition: A triangle is isosceles if two if its sides are equal.

We want to prove the following properties of isosceles triangles.

Theorem: Let ABC be an isosceles triangle with AB = AC.  Let M denote the midpoint of BC (i.e., M is the point on BC for which MB = MC).  Then

a)      Triangle ABM is congruent to triangle ACM.

b)      Angle ABC = Angle ACB (base angles are equal)

c)      Angle AMB = Angle AMC = right angle.

d)      Angle BAM = angle CAM

Corollary: Consequently, from these facts and the definitions:

Ray AM is the angle bisector of angle BAC.

Line AM is the altitude of triangle ABC through A.

Line AM is the perpendicular bisector of B

Segment AM is the median of triangle ABC through A.

Proof #1 of Theorem (after B&B)

Let the angle bisector of BAC intersect segment BC at point D.  

Since ray AD is the angle bisector, angle BAD = angle CAD.  

The segment AD = AD = itself.

Also, AB = AC since the triangle is isosceles.

Thus, triangle BAD is congruent to CAD by SAS (side-angle-side).

This means that triangle BAD = triangle CAD, and corresponding sides and angles are equal, namely:

DB = DC,

angle ABD = angle ACD,

angle ADB = angle ADC.

(Proof of a).  Since DB = DC, this means D = M by definition of the midpoint.  Thus triangle ABM = triangle ACM.

(Proof of b) Since angle ABD = angle ABC (same angle) and also angle ACD = angle ACB, this implies angle ABC = angle ACB.

(Proof of c) From congruence of triangles, angle AMB = angle AMC.  But by addition of angles, angle AMB + angle AMC = straight angle = 180 degrees.  Thus 2 angle AMB = straight angle and angle AMB = right angle.

(Proof of d) Since D = M, the congruence angle BAM = angle CAM follows from the definition of D.  (These are also corresponding angles in congruent triangles ABM and ACM.)

QED*

*Note:  There is one point of this proof that needs a more careful “protractor axiom”.  When we constructed the angle bisector of BAC, we assumed that this ray intersects segment BC.  This can’t be quite deduced from the B&B form of the axioms.  One of the axioms needs a little strengthening.

The other statements are immediate consequence of these relations and the definitions of angle bisector, altitude, perpendicular bisector, and median.  (Look them up!)

Definition:  We will call the special line AM the line of symmetry of the isosceles triangle.  Thus we can construct AM as the line through A and the midpoint, or the angle bisector, or altitude or perpendicular bisector of BC. Shortly we will give a general definition of line of symmetry that applies to many kinds of figure.

Proof #2 (This is a slick use of SAS, not presented Monday.  We may discuss in class Wednesday.)

The hypothesis of the theorem is that AB = AC.  Also, AC = AB (!) and angle BAC = angle CAB (same angle).  Thus triangle BAC is congruent to triangle BAC by SAS.

The corresponding angles and sides are equal, so the base angle ABC = angle ACB.

Let M be the midpoint of BC.  By definition of midpoint, MB = MC. Also the equality of base angles gives angle ABM = angle ABC = angle ACB = angle ACM.  Since we already are given BA = CA, this means that triangle ABM = triangle ACM by SAS.

From these congruent triangles then we conclude as before:

Angle BAM = angle CAM (so ray AM is the bisector of angle BAC)

Angle AMB = angle AMC = right angle (so line MA is the perpendicular bisector of  BC and also the altitude of ABC through A)

QED

Faulty Proof #3.  Can you find the hole in this proof?)

In triangle ABC, AB = AC.  Let M be the midpoint and MA be the perpendicular bisector of BC.

Then angle BMA = angle CMA = right angle, since MA is perpendicular bisector.  

MB = MC by definition of midpoint. (M is midpoint since MA is perpendicular bisector.)

AM = AM (self).

So triangle AMB = triangle AMC by SAS.

Then the other equal angles ABC = ACB and angle BAM = angle CAM follow from corresponding parts of congruent triangles.  And the rest is as before.

QED??

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