1) (3.14) The total scores on the Medical College Admission Test (MCAT) follow a

Question

1) (3.14) The total scores on the Medical College Admission Test (MCAT) follow a Normal distribution with mean 24.4 and standard deviation 6.2 .

What are the median and the first and third quartiles of the MCAT scores?

(Use software to calculate z)

Q1 (±0.001) =

Median (±0.001) =

Q3 (±0.001) =

2) Suppose the length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 13 days. 95% of all pregnancies last between....?

3) Find the proportion of observations (±0.0001) from a standard Normal distribution that falls in each of the following regions. In each case, sketch a standard Normal curve and shade the area representing the region.

(a)z2.34:

(b)z2.34:

(c)z>1.74:

(d)2.34<z<1.74:

Explanation / Answer

1.

FOR Q1:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.25      
          
Then, using table or technology,          
          
z =    -0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    24.4      
z = the critical z score =    -0.67448975      
s = standard deviation =    6.2      
          
Then          
          
x = critical value =    20.21816355   [ANSWER, Q1]

****************************

FOR MEDIAN:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.5      
          
Then, using table or technology,          
          
z =    0      
          
As x = u + z * s,          
          
where          
          
u = mean =    24.4      
z = the critical z score =    0      
s = standard deviation =    6.2      
          
Then          
          
x = critical value =    24.4   [ANSWER, MEDIAN]

******************************

FOR Q3:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.75      
          
Then, using table or technology,          
          
z =    0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    24.4      
z = the critical z score =    0.67448975      
s = standard deviation =    6.2      
          
Then          
          
x = critical value =    28.58183645   [ANSWER, Q3]  
  

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