# desperate help on s1!! help greatly needed!!Watch

#1
hello
i need help on s1, it involves simultaneous equations and im finding it hard. please help!! thank you!!

a company produces electronic componenets which have life spans that are normally distributed. only 1% of the components have a life span less than 3500 hours and 2.5% have a life span greater than 5500 hours.

Determine themean and standard deviation of the life spans of the components.
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12 years ago
#2
When p=1%, z=-2.3263; when p=2.5% z=1.96 (the z values come from the second normal distribution table).
So:
-2.3263=(3500-m)/s rearrange to get m=3500+2.3263s
1.96=(5500-m)/s rearrange to get m=5500-1.96s
5500-1.96s=3500+2.3263s rearrange to get s=2000/4.2863=466.6
m=3500+(2.3263*466.6)=4585.5
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#3
thank u!
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