Question

The numbers of minutes that travellers arrives at the airport before their scheduled departure are approximately normally distributed with mean 80.5 and standard deviation 9.9. what proportion of travellers arrive at the airport more than 77.4 minutes before their departure? write only a number as your answer. round to 4 decimal places. do not write as a percentage.

Answer 1

Solution :

Given that ,

mean = $\mu$ = 80.5

standard deviation = $\sigma$ = 9.9

P(x > 77.4) = 1 - P(x <77.4 )

= 1 - P[(x - $\mu$ ) / $\sigma$ < (77.4 - 80.5) /9.9 ]

= 1 - P(z < -0.31)

Using z table,

= 1 -0.3783

=0.6217