Question

Terri Vogel, an amateur motorcycle racer, averages 129.71 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)

• a. In words, define the random variable X. X=
• b. X~
• c. Find the percent of her laps that are completed in less than 130 seconds.
• d. The fastest 3% of her laps are under _______ .
• e. The middle 80% of her laps are from _______ seconds to _______ seconds.

Answer 1

a)

X = Obeserving race time of Terri Vogel to complete 2.5 mile lap.

b)

X ~ N ( 129.71 , 2.28)

c)

X ~ N ( µ = 129.71 , σ = 2.28 )
We convert this to standard normal as
P ( X < x ) = P ( Z < ( X - µ ) / σ )
P ( ( X < 130 ) = P ( Z < 130 - 129.71 ) / 2.28 )
= P ( Z < 0.13 )
P ( X < 130 ) = 0.5517
= 55.17%

d)

P ( X < x ) = 3% = 0.03
To find the value of x
Looking for the probability 0.03 in standard normal table to calculate critical value Z = -1.8808
Z = ( X - µ ) / σ
-1.8808 = ( X - 129.71 ) / 2.28
X = 125.42

e)

P ( a < X < b ) = 0.8
Dividing the area 0.8 in two parts we get 0.8/2 = 0.4
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.4
Area above the mean is b = 0.5 + 0.4
Looking for the probability 0.1 in standard normal table to calculate critical value Z = -1.2816
Looking for the probability 0.9 in standard normal table to calculate critical value Z = 1.2816
Z = ( X - µ ) / σ
-1.2816 = ( X - 129.71 ) / 2.28
a = 126.79
1.2816 = ( X - 129.71 ) / 2.28
b = 132.63

The middle 80% of her laps are from 126.79 seconds to 132.63 seconds