Question

1.The heights of women aged 20 to 29 follow approximately the N(64, 2.57) distribution. Men the same age have heights distributed as N(69.3, 2.8). What percent of young men are shorter than the mean height of young women?

2. Changing the mean and standard deviation of a Normal distribution by a moderate amount can greatly change the percent of observation in the tails. Suppose that a college is looking for applicants with SAT math scores 760 and above.

(a) In 2007, the scores of men on the math SAT followed the N(533, 116) distribution. What percent of men scored 760 or better?
(b) Women's SAT math scores that year had the N(499, 110) distribution. What percent of women scored 760 or better?

3. Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=101μ=101 and σ=24σ=24.

(a) What proportion of children aged 13 to 15 years old have scores on this test above 87 ? (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 25 percent of the distribution.
Answer:

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

Answer 1

1)

µ =    69.3          
σ =    2.8          
              
P( X < 64   ) = P( (X-µ)/σ ≤ (64-69.3) /2.8)      
=P(Z < -1.893   ) = 0.0292 or 2.92% (answer)

excel formula for probability from z score is =NORMSDIST(Z)

2)

a)

µ =    533                  
σ =    116                  
                      
P ( X ≥   760   ) = P( (X-µ)/σ ≥ (760-533) / 116)              
= P(Z ≥   1.957   ) = P( Z <   -1.957   ) =    0.0252 or 2.52% (answer)

b)

µ =    499                  
σ =    110                  
                      
P ( X ≥   760   ) = P( (X-µ)/σ ≥ (760-499) / 110)              
= P(Z ≥   2.373   ) = P( Z <   -2.373   ) =    0.0088 or 0.88% (answer)