Question

A safety light is designed so that the time between flashes are normally distributed with a mean of 2.50 s and a standard deviation of 0.60 s

Find the probability that an individual time is greater than 3.00 s

Find the probability that the mean for 50 randomly selected times is greater than 3.00 s.

Given that the light is intended to help people see an​ obstruction, which result is more relevant for assessing the safety of the​ light?

Answer 1

Solution :

Given that ,

mean = = 2.50

standard deviation = = 0.60

P(x > 3.00 ) = 1 - P ( x < 3.00 )

= 1- P[(x - ) / < ( 3.00 - 2.50) / 0.60 ]

= 1 - P ( z < 0.83 )

Using z table,

= 1 - 0.7967

= 0.2033

Probability = 0.2033

n = 50

= 2.50

= / n = 0.60 / 50 = 0.0849

P( >3.00) = 1 - P( < 3.00)

= 1 - P[( - ) / < ( 3.00 2050 ) / 0.0849 ]

= 1 - P ( z < 5.89 )

Using z table,    

= 1 - 1.0000

= 0.0000

Probability = 0.0000