Question

 

In a recent race, with mean 210 min, and standard deviation 25 min, the finish times formed a normal distribution:

1. Find the z-score of Jose, who finished in 190 minutes.
2. Find the z-score of Maria, who finished in 270 minutes.
3. What is the probability that a racer would finish the race in less than 180 minutes?
4. What is the probability that a racer would finish the race between 190 and 225 minutes?

Answer 1

Solution :

mean = = 210

standard deviation = = 25

a ) X = 190

Using z-score formula,

z = X -   /

= 190 - 210 / 25

= - 20 / 25

= - 0.8

The z-score = - 0.8

b ) X = 270

Using z-score formula,

z = X -   /

= 270 - 210 / 25

= 60 / 25

= 2.4

The z-score = 2.4

c ) P( x < 180 )

P ( x -  / ) < ( 180 - 210 / 25)

P ( z < - 30 / 25 )

P ( z < - 1. 2 )

Using z table

= 0.1151

Probability = 0.1151

d ) P ( 190 < x < 225 )

P ( 190 - 210 / 25) < ( x -  / ) < ( 225 - 210 / 25)

P ( - 20 / 25 < z < 15 / 25 )

P (- 0.8 < z < 0.6 )

P (z < 0.6 ) - p ( z < - 0.8 )

Using z table

= 0.7257 - 0.2119

= 0.5138

Probability = 0.5138