# guptaneeraj8888

## Activity

• 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? 1 year, 8 months ago

• @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.

• According to a manager it takes an average weekday commute of 39 minutes with a Standard Deviation of 7 minutes for the employees to get to work when they use their personal vehicles for their office commute while management set a policy of not more than 40 minutes for their daily one-way commute. A survey conducted one day on 70 employees showed…[Read more]

• 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 year, 8 months ago

• 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