Question

Terri Vogel, an amateur motorcycle racer, averages 129.11 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-NG b. Find the proportion of her laps that are completed between 125.77 and 128.66 seconds. c. The fastest 3% of laps are under seconds. d. The middle 80% of her laps are from seconds to seconds.

Answer 1

Part a)

X ~ N ( µ = 129.11 , σ = 2.28 )

part b)

P ( 125.77 < X < 128.66 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 125.77 - 129.11 ) / 2.28
Z = -1.4649
Z = ( 128.66 - 129.11 ) / 2.28
Z = -0.1974
P ( -1.46 < Z < -0.2 )
P ( 125.77 < X < 128.66 ) = P ( Z < -0.2 ) - P ( Z < -1.46 )
P ( 125.77 < X < 128.66 ) = 0.4218 - 0.0715
P ( 125.77 < X < 128.66 ) = 0.3503

Part c)

X ~ N ( µ = 129.11 , σ = 2.28 )
P ( X < x ) = 3% = 0.03
To find the value of x
Looking for the probability 0.03 in standard normal table to calculate Z score = -1.8808
Z = ( X - µ ) / σ
-1.8808 = ( X - 129.11 ) / 2.28
X = 124.8218 ≈ 124.82 seconds
P ( X < 124.8218 ) = 0.03

Part d)

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 Z score = -1.2816
Looking for the probability 0.9 in standard normal table to calculate Z score = 1.2816
Z = ( X - µ ) / σ
-1.2816 = ( X - 129.11 ) / 2.28
a = 126.188 ≈ 126.19 seconds
1.2816 = ( X - 129.11 ) / 2.28
b = 132.032 ≈ 132.03 seconds
P ( 126.19 < X < 132.03 ) = 0.8