Question

1. The average price for a new car is $18,000 and the standard deviation is $1,500. Calculate the z scores for the following individual scores. SHOW YOUR WORK.If X = 19,000, z = _

2. Using z-scores. SHOW YOUR WORK

The percent of area beyond z = 1.44 is

3. Using z-scores. SHOW YOUR WORK

).The percent of area between the z = 1.40 and z = 1.49 is ________.

4. Using z-scores. SHOW YOUR WORK

The percent of TOTAL cases that have a z-score less than or equal to 1.46 is ________.

5. The mean for IQ is 100 and the sd is 15. Answer the following questions using that information. SHOW YOUR WORK

20% of observations fall beyond a certain score, what is the z-score? __________.

6. The mean for IQ is 100 and the sd is 15. Answer the following questions using that information. SHOW YOUR WORK

Take the z-score from the previous question and identify the IQ score at the 80th percentile __________.

Answer 1

1)  
Z = (X-mean)/sd  
z =   =(19000-18000)/1500
z =   0.6667
  
2)  
P(Z<1.44) = ?  
= NORMSDIST(1.44)  
= 0.9251  
  
3)  
P(1.4 < Z < 1.49)  
= P(Z<1.49)-P(Z<1.4)  
= NORMSDIST(1.49)-NORMSDIST(1.4)  
= 0.0126  
  
4)  
P(Z<=1.46)  
= NORMSDIST(1.46)  
= 0.9279  
  
5)  
P(Z<z) = 20%  
z =    =NORMSINV(0.2)
z =    -0.8416
(X-mean)/sd =    -0.8416
X =    =-0.8416*15+100
X =    87.376
  
6)  
P(Z<z) =    80%
Z =    =NORMSINV(0.8)
Z =    0.841621234
(X-mean)/sd =    0.841621234
X =    =0.8416*15+100
X =    112.624