Question

Terri Vogel, an amateur motorcycle racer, averages 129.99 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.25 seconds . The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps. (Source: log book of Terri Vogel) Let X be the number of seconds for a randomly selected lap. Round all answers to 4 decimal places where possible. a. What is the distribution of X? X ~ N( , ) b. Find the proportion of her laps that are completed between 127.18 and 129.89 seconds. c. The fastest 2% of laps are under seconds. d. The middle 40% of her laps are from seconds to seconds.

Answer 1

Solution :

Given that ,

mean = = 129.99

standard deviation = = 2.25

a) The distribution of x is normal X N ( 129.99, 2.25 )

b) P(127.18 < x < 129.89) = P[(127.18 - 129.99)/ 2.25) < (x - ) /  < (129.89 - 129.99) / 2.25 ) ]

= P(-1.25 < z < -0.04)

= P(z < -0.04) - P(z < -1.25)

Using z table,

= 0.4840 - 0.1056

= 0.3784

c) Using standard normal table,

P(Z > z) = 2%

= 1 - P(Z < z) = 0.02  

= P(Z < z) = 1 - 0.02

= P(Z < z ) = 0.98

= P(Z < 2.05 ) = 0.98  

z = 2.05

Using z-score formula,

x = z * +

x = 2.05 * 2.25 + 129.99

x = 134.60 seconds.

d) Using standard normal table,

P( -z < Z < z) = 40%

= P(Z < z) - P(Z <-z ) = 0.40

= 2P(Z < z) - 1 = 0.40

= 2P(Z < z) = 1 + 0.40

= P(Z < z) = 1.40 / 2

= P(Z < z) = 0.70

= P(Z < 0.52) = 0.70

= z  ± 0.52

Using z-score formula,

x = z * +

x = -0.52 * 2.25 + 129.99

x = 128.82 seconds

Using z-score formula,

x = z * +

x = 0.52 * 2.25 + 129.99

x = 131.16 seconds

The middle 40% are from 128.82 seconds  to 131.16 seconds.