Question

Terri Vogel, an amateur motorcycle racer, averages 129.96 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.28 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 126.48 and 127.95 seconds.  

c. The fastest 2% of laps are under  seconds.

d. The middle 50% of her laps are from  seconds to  seconds.  

Answer 1

Solution :

a.

X $\sim$ N (129.96 , 2.28)

b.

P(126.48 < x < 127.95) = P[(126.48 - 129.96)/ 2.28) < (x - $\mu$ ) /$\sigma$  < (127.95 - 129.96) / 2.28) ]

= P(-1.53 < z < -0.88)

= P(z < -0.88) - P(z < -1.53)

= 0.1894 - 0.063

= 0.1264

proportion = 0.1264

c.

Using standard normal table ,

P(Z > z) = 2%

1 - P(Z < z) = 0.02

P(Z < z) = 1 - 0.02

P(Z < 2.05) = 0.98

z = 2.05

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 2.05 * 2.28 + 129.96 = 134.6340

134.6340  seconds.  

d.

Middle 50% as the to z values are -0.674 and 0.674

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -0.674 * 2.28 + 129.96 = 128.4233

and

x = 0.674 * 2.28 + 129.96 = 131.4968

128.4233 seconds to 131.4968 seconds.