Question

Terri Vogel, an amateur motorcycle racer, averages 129.63 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 127.41 and 129.71 seconds.

c. The fastest 3% of laps are under  seconds.

d. The middle 70% of her laps are from  seconds to  seconds.

Answer 1

Solution :

Given that ,

mean = = 129.63

standard deviation = = 2.28

a.

X N (129.63 , 2.28)

b.

P(127.41 < x < 129.71) = P[(127.41 - 129.63)/ 2.28) < (x - ) /  < (129.71 - 129.63) / 2.28) ]

= P(-0.97 < z < 0.04)

= P(z < 0.04) - P(z < -0.97)

= 0.516 - 0.166

= 0.3500

proportion = 0.3500

c.

Using standard normal table ,

P(Z > z) = 3%

1 - P(Z < z) = 0.03

P(Z < z) = 1 - 0.03

P(Z < 1.88) = 0.97

z = 1.88

Using z-score formula,

x = z * +

x = 1.88 * 2.28 + 129.63 = 133.92

The fastest 3% of laps are under 133.92 seconds.

d.

Middle 70% as the to z values are -1.036 and 1.036

Using z-score formula,

x = z * +

x = -1.036 * 2.28 + 129.63 = 127.27

and

x = 1.036 * 2.28 + 129.63 = 131.99

The middle 70% of her laps are from 127.27 seconds to 131.99 seconds.