Question

(1 point) Almost all medical schools in the United States require applicants to take the Medical College Admission Test (MCAT). On one exam, the scores of all applicants on the biological sciences part of the MCAT were approximately Normal with mean 9.4 and standard deviation 2.2. For applicants who actually entered medical school, the mean score was 10.2 and the standard deviation was 1.7.

(a) What percent of all applicants had scores higher than 12?
ANSWER:   

%

(b) What percent of those who entered medical school had scores between 10 and 11?
ANSWER:   

%

(1 point) The temperature at any random location in a kiln used in the manufacture of bricks is normally distributed with a mean of 1025 and a standard deviation of 40 degrees.
If bricks are fired at a temperature above 1140, they will crack and must be disposed of. If the bricks are placed randomly throughout the kiln, the proportion of bricks that crack during the firing process is closest to

(Note: Please enter your answer in decimal form. For example, 5.02% should be entered as 0.0502.)

(1 point) Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=106μ=106 and σ=24σ=24.

(a) What proportion of children aged 13 to 15 years old have scores on this test above 96 ? (NOTE: Please enter your answer in decimal form. For example, 45.23% should be entered as 0.4523.)
Answer:

(b) Enter the score which marks the lowest 30 percent of the distribution.
Answer:

(c) Enter the score which marks the highest 10 percent of the distribution.
Answer:

Answer 1

1)

a)

P ( X ≥   12   ) = P( (X-µ)/σ ≥ (12-9.4) / 2.2)              
= P(Z ≥   1.182   ) = P( Z <   -1.182   ) =    11.86 % (answer)

b)

µ =    10.2                                  
σ =    1.7                                  
we need to calculate probability for ,                                      
P (   10   < X <   11   )                      
=P( (10-10.2)/1.7 < (X-µ)/σ < (11-10.2)/1.7 )                                      
                                      
P (    -0.118   < Z <    0.471   )                       
= P ( Z <    0.471   ) - P ( Z <   -0.118   ) =    0.6810   -    0.4532   = 22.79% (answer)
====================

2)

µ =    1025                  
σ =    40                  
                      
P ( X ≥   1140   ) = P( (X-µ)/σ ≥ (1140-1025) / 40)              
= P(Z ≥   2.875   ) = P( Z <   -2.875   ) =    0.0020   (answer)
================

3)

a)

P ( X ≥   96   ) = P( (X-µ)/σ ≥ (96-106) / 24)              
= P(Z ≥   -0.417   ) = P( Z <   0.417   ) =    0.6615   (answer)

b)

Z value at    0.3   =   -0.524   (excel formula =NORMSINV(   0.3   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -0.524   *   24   +   106  
X   =   93.4144   (answer)          

c)

Z value at    0.9   =   1.282   (excel formula =NORMSINV(   0.9   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.282   *   24   +   106  
X   =   136.7572   (answer)