Question

4. Convert the following x-scores to z-scores:

                a) 8                                                                                        b) 21.5

5. Convert the following z-scores to x-scores:

                a) .2667

6.Find the area under the curve:

a) to the left of z = 1.15                                                                  b) to the left of z = -0.24

c) to the right of z = 1.06                                                           d) between z = 1.25 and z = -1.

7.A survey indicates that people use their computers an average of 2.4 years before upgrading to a new machine. The standard deviation is 0.5 year. If a computer owner is selected at random, find the probability that he or she will use it for less than 2 years before upgrading. Assume that the variable x is normally distributed.

8Scores for a civil service exam are normally distributed, with a mean of 75 and a standard deviation of 6.5. To be eligible for civil service employment, you must score in the top 5%. What is the lowest score you can earn and still be eligible for employment?

9.The length of time employees have worked at a corporation is normally distributed with a mean of 11.2 years and a standard deviation of 2.1 years. In a company cutback, the lowest 10% in seniority are laid off. What is the maximum length of time an employee could have worked and still be laid off?         

Answer 1

6)

a) =P(Z ≤   1.15   ) =   0.8749   (answer)

b) =P(Z ≤   -0.24   ) =   0.4052   (answer)

c) = P(Z ≥   1.06   ) = P( Z <   -1.060   ) =    0.14457   (answer)

d) P (    -1.000   < Z <    1.250   )                       
= P ( Z <    1.250   ) - P ( Z <   -1.00   ) =    0.8944   -    0.1587   =    0.7357   (answer)

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

7)

µ =    2.4          
σ =    0.5          
              
P( X ≤    2   ) = P( (X-µ)/σ ≤ (2-2.4) /0.5)      
=P(Z ≤   -0.80   ) =   0.2119   (answer)

8)

µ=   75                  
σ =    6.5                  
P(X≤x) =   0.95                  
                      
Z value at    0.95   =   1.6449   (excel formula =NORMSINV(   0.95   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.645   *   6.5   +   75  
X   =   85.69   (answer)          

9)

µ=   11.2                  
σ =    2.1                  
P(X≤x) =   0.1                  
                      
Z value at    0.1   =   -1.2816   (excel formula =NORMSINV(   0.1   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -1.282   *   2.1   +   11.2  
X   =   8.51   (answer)