Question

For the following data, construct a frequency distribution with six classes.

57	23	35	18	21
26	51	47	29	21
46	43	29	23	39
50	41	19	36	28
31	42	43	29	18
28	46	33	28	20

Class Interval	Frequencies
16 - under 23	enter a frequency
23 - under 30	enter a frequency
30 - under 37	enter a frequency
37 - under 44	enter a frequency
44 - under 51	enter a frequency
51 - under 58	enter a frequency
TOTAL	enter a total for the column

2. can you look at this and help me

According to a Human Resources report, a worker in the industrial countries spends on average 419 minutes a day on the job. Suppose the standard deviation of time spent on the job is 30 minutes.

a. If the distribution of time spent on the job is approximately bell shaped, between what two times would 68% of the figures be?
enter the lower limit for the interval where 68% of the values would fall  to enter the upper limit for the interval where 68% of the values would fall   

b. If the distribution of time spent on the job is approximately bell shaped, between what two times would 95% of the figures be?
enter the lower limit for the interval where 95% of the values would fall  to enter the upper limit for the interval where 95% of the values would fall   

c. If the distribution of time spent on the job is approximately bell shaped, between what two times would 99.7% of the figures be?
enter the lower limit for the interval where 99.7% of the values would fall  to enter the upper limit for the interval where 99.7% of the values would fall   

d. If the shape of the distribution of times is unknown, approximately what percentage of the times would be between 361 and 477 minutes?
enter percentages rounded to 1 decimal place _________ %  (Round the intermediate values to 3 decimal places. Round your answer to 1 decimal place.)

e. Suppose a worker spent 400 minutes on the job. What would that worker’s z score be, and what would it tell the researcher?
z score = enter the z score rounded to 3 decimal places  (Round your answer to 3 decimal places.)

This worker is in the lower half of workers but within select the distance from the mean one standard deviation of the mean.

Answer 1

1)

Class Interval	Frequencies
16 - under 23	6
23 - under 30	9
30 - under 37	4
37 - under 44	5
44 - under 51	4
51 - under 58	2
TOTAL	30

2)

A)

µ =    419                      
σ =    30                      
proportion=   0.6800                      
proportion left    0.3200   is equally distributed both left and right side of normal curve                   
z value at   0.16   = ±   0.994   (excel formula =NORMSINV(   0.32   / 2 ) )  
                          
z = ( x - µ ) / σ                          
so, X = z σ + µ =                          
X1 =   -0.994   *   30   +   419   =   389.17
X2 =   0.994   *   30   +   419   =   448.83

...........

B)

µ =    419                      
σ =    30                      
proportion=   0.9500                      
proportion left    0.0500   is equally distributed both left and right side of normal curve                   
z value at   0.025   = ±   1.960   (excel formula =NORMSINV(   0.05   / 2 ) )  
                          
z = ( x - µ ) / σ                          
so, X = z σ + µ =                          
X1 =   -1.960   *   30   +   419   =   360.20
X2 =   1.960   *   30   +   419   =   477.80
..

C)

µ =    419                      
σ =    30                      
proportion=   0.9970                      
proportion left    0.0030   is equally distributed both left and right side of normal curve                   
z value at   0.0015   = ±   2.968   (excel formula =NORMSINV(   0.00   / 2 ) )  
                          
z = ( x - µ ) / σ                          
so, X = z σ + µ =                          
X1 =   -2.968   *   30   +   419   =   330
X2 =   2.968   *   30   +   419   =   508

..............

D)

µ =    419                              
σ =    30                              
we need to calculate probability for ,                                  
P (   361   < X <   477   )                  
=P( (361-419)/30 < (X-µ)/σ < (477-419)/30 )                                  
                                  
P (    -1.933   < Z <    1.933   )                   
= P ( Z <    1.933   ) - P ( Z <   -1.933   ) =    0.9734   -    0.0266   =    0.9468

= 94.7%

......

E)

µ=   419
σ=   30
X=   400
Z=(X-µ)/σ=   (400-419)/30)=       -0.633

....

THANKS

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