Question

Problem 2 Kasi Skyrider, an amateur motorcyde racer, averages 135 seconds per lap with a standard deviation of 3.4 seconds. The distribution of her lap times is normally distributed. We are interested in one of her randomly selected laps. a. (2%) X ~ trick and b. (3%) What is the probability that her laps are completed in less than 130 seconds? (3%) The fastest 3% of her laps are completed in under seconds d. (3%) The middle 80% of her laps are completed in between seconds and econds

Answer 1

Part a)

X ~ N ( µ = 135 , σ = 3.4 )

Part b)
P ( X < 130 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 130 - 135 ) / 3.4
Z = -1.4706
P ( ( X - µ ) / σ ) < ( 130 - 135 ) / 3.4 )
P ( X < 130 ) = P ( Z < -1.4706 )
P ( X < 130 ) = 0.0707

Part c)

X ~ N ( µ = 135 , σ = 3.4 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.03 = 0.97
To find the value of x
Looking for the probability 0.97 in standard normal table to calculate Z score = 1.8808
Z = ( X - µ ) / σ
1.8808 = ( X - 135 ) / 3.4
X = 141.3947
P ( X > 141.3947 ) = 0.03

Part d)

X ~ N ( µ = 135 , σ = 3.4 )
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 - 135 ) / 3.4
a = 130.6426
1.2816 = ( X - 135 ) / 3.4
b = 139.3574
P ( 130.6426 < X < 139.3574 ) = 0.8