The Student Room Group

S1 May 2002 question!

A company produces electronic components 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.

a) Determine the mean and standard deviation of the life spans of the components.

Please explain step by step.
Thanks

Reply 1

Mohit_C
A company produces electronic components 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.

a) Determine the mean and standard deviation of the life spans of the components.

Please explain step by step.
Thanks



Let X be the life span in 1000's of hours.
X~N(μ,σ2)

so P(X< 3.5) = 0.01

P(Z< (3.5 - &#956:wink:/&#963:wink: = 0.01 (1)

P(X>5.5) = 0.02
P(Z>(5.5 - &#956:wink:/&#963:wink: = 1 - P(Z&#8804;(5.5 - &#956:wink:/&#963:wink: = 0.02 (2)

use ur normal distribution tables to work solve (1) and (2) simultaneously to get &#956; and &#963;

Reply 2

Hey has anyone got the mark scheme for S1 may 2002

i could really do with it :smile:

Thanks