I need help with finding percentage after solving z score.

Statistics
Tutor: None Selected Time limit: 1 Day

 mean=35,000 km

 standard deviation=1500 km

What percent of tires will have a life that exceeds 26,000 km?

If 2000 tires are purchased, how many tires would you expect to last more than 28,000 km?

Sep 25th, 2015

Thank you for the opportunity to help you with your question!

1. We need P(x>26000)

first, transform x to z- score: z=(26000-35000)/15000=-.6

P(z>-.6)= 1-P(z<-.6)=1-.2743=.7257=72.57%

2. P(x>28000)=P(z>(28-35)/15)=-.47

P(z>-.47)=1-P(z<-.47)=1-.3192=.6808 or 68.08%

68.08% of 2000=1362 would have life exceeding 28,000 km.


Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 26th, 2015

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Sep 25th, 2015
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