Question

Terri Vogel, an amateur motorcycle racer, averages 129.73 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.25 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 129.38 and 131.04 seconds. ____

c. The fastest 4% of laps are under ____ seconds.

d. The middle 60% of her laps are from ____ seconds to ____ seconds.

Answer 1

a)X ~ N(129.73 , 2.25)

b)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	129.73
std deviation   =σ=	2.2500

probability =

P(129.38<X<131.04)

=

P(-0.16<Z<0.58)=

0.7190-0.4364=

0.2826

c)

for 4th percentile critical value of z=		-1.75
therefore corresponding value=mean+z*std deviation=			125.79 seconds

d)

The middle 60% of her laps are from _127.84_ seconds to _131.62_ seconds