Question

Terri Vogel, an amateur motorcycle racer, averages 129.3 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 129.9 and 131.44 seconds. ____   

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

d. The middle 80% of her laps are from ___ seconds to ____seconds.

Answer 1

a) X~ N(129.3 ,2.28)

b)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	129.3
std deviation   =σ=	2.2800

probability =

P(129.9<X<131.44)

=

P(0.26<Z<0.94)=

0.8264-0.6026=

0.2238

c)

for 3th percentile critical value of z=		-1.88
therefore corresponding value=mean+z*std deviation=			125.01 seconds

d)

for middle 80% CI ; critical z = -/+ 1.28

therefore

he middle 80% of her laps are from 126.38 seconds to 132.22__seconds.