Kevin uses his mobile phone for X minutes each day. X is a random variable which may be modelled by a normal distribution. Given that 80% if X lie on a symmetrical interval of 17.7476 minutes and 38.2524 minutes. i) Determine the mean and standard deviation. ii) Hence, find the probability that on a particular day, Kevin uses his mobile phone between 10 and 20 minutes. iii) Find the probability that on 7 randomly selected days the mean time Kevin spends on his mobile phone is at least 30 minutes. iv) Hence, find the interval of X that on 7 randomly selected days Kevin spends on his mobile phone when the probability is 0.8
Kevin uses his mobile phone for X minutes each day. X is a random variable which may be modelled by a normal distribution. Given that 80% if X lie on a symmetrical interval of 17.7476 minutes and 38.2524 minutes. i) Determine the mean and standard deviation. ii) Hence, find the probability that on a particular day, Kevin uses his mobile phone between 10 and 20 minutes. iii) Find the probability that on 7 randomly selected days the mean time Kevin spends on his mobile phone is at least 30 minutes. iv) Hence, find the interval of X that on 7 randomly selected days Kevin spends on his mobile phone when the probability is 0.8
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Kevin uses his mobile phone for X minutes each day. X is a random variable which may be modelled by a normal distribution . Given that 80% if X lie on a symmetrical interval of 17.7476 minutes and 38.2524 minutes. i) Determine the mean and standard deviation. ii) Hence, find the probability that on a particular day, Kevin uses his mobile phone between 10 and 20 minutes. iii) Find the probability that on 7 randomly selected days the mean time Kevin spends on his mobile phone is at least 30 minutes. iv) Hence, find the interval of X that on 7 randomly selected days Kevin spends on his mobile phone when the probability is 0.8.
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