# guptaneeraj8888

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Viewing 4 posts - 1 through 4 (of 4 total)
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• #236394

guptaneeraj8888
Participant

I have calculated the Z score as it represent the probability by considering X=70
Z= 70-39/7= 4.42
Now, in order to calculate variance for personal transportation I have done 1-Zscore= 1-4.42=3.42.
So we should deduct 3.42 in order to increase their chances to come on time using personal transportation.
Am i correct?

0
#236390

guptaneeraj8888
Participant

@KatieBarry this is not a home work question. I am not asking for a complete solution but a general approach. May be my understanding of concept is weak but that not make my effort less. I have tried several approach which includes following :

I have tried by calculating the z score for the people using personal transport by assuming n= 70.

Also tried by plotting all the data on a normal distribution curve.

Is my approach is correct or I need to go through a different route?

0
#236140

guptaneeraj8888
Participant

Gotcha!!!
For overall process DPMO : 10+17/[(500*15) + (1000*5)] *1000000 = 27/12500 *1000000 = .00216* 1000000 =2160.

Thanks Aaron.
Will asking for further doubt ;-)

1
#236137

guptaneeraj8888
Participant

I tried but got confused with the defectives. Following is the steps which I have done:
For servivce A:
DPMO:10/(500*15)*1000000= 1333.33

For Service B:
DPMO: 17/(1000*5)*1000000=3400

For overall process: 1333.33+3400=4733.33

Please suggest am I doing it right.
PS: This a practice test question and I got stuck because it answer is
2160.

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