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:
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