Question

Suppose the average speeds of passenger trains traveling from Newark, New Jersey, to Philadelphia, Pennsylvania, are normally distributed, with a mean average speed of 88 miles per hour and a standard deviation of 6.4 miles per hour.

(a) What is the probability that a train will average less than 72 miles per hour?

(b) What is the probability that a train will average more than 80 miles per hour?

(c) What is the probability that a train will average between 91 and 99 miles per hour?

(Round the values of z to 2 decimal places. Round your answers to 4 decimal places.)

(a) P(x < 72)=???

(b) P(x > 80)=???

(c) P(91 ≤ x ≤ 99)=???

Answer 1

Solution :

Given that ,

mean = = 88

standard deviation = = 6.4

a) P(x < 72 ) = P[(x - ) / < (72 - 88) / 6.4]

= P(z < -2.50)

Using z table,

= 0.0062

b) P(x > 80) = 1 - p( x< 80)

=1- p P[(x - ) / < (80 - 88) / 6.4 ]

=1- P(z < -1.25)

Using z table,

= 1 - 0.1056

= 0.8944

c) P(91 x 99)

= P[(91 - 88 / 6.4) (x - ) / ( 99 - 88 / 6.4) ]

= P(0.47 z 1.72)

= P(z 1.72) - P(z 0.47)

Using z table,  

= 0.9573 - 0.6808

= 0.2765