Question

b. The length of time, L hours, that a laptop will work before it needs charging is normally distributed with mean of 50 hours and a standard deviation of 15 hours Find the percentage of the laptop that will work between 56 and 60 hours. [3 marks] If 10 % of the laptop still working without charging for more than h il hours, find the minimum value of h. [4 marks] Afiqa is about to go on a 6-hours journey. Given that it is 57-hours since Afiqa last charged her laptop, find the probability that her laptop will not be charging before her journey is completed. [3 marks] 5

Answer 1

9 M = 50,5 = 15,7 -- *-^ - *SO OP(56<x560) - Plés< zales) = P(0.4<zs1033) = P(Z <1.33) - Plż<0.4) -0.9082-0.6554 = P(x>)=0.10 » PLZ 6-50 ):0.10 Piza h-50 ):0.90 also pizs1,28):0.90 (-50)/15 - 01.99 ha in P(x63\x757) = P(x763) - Plz> 0.87) - 4-P (2<0.87) - 1-0.8078 i-plz<0.42) -I-0.6808 0.6021)