Question

Terri Vogel, an amateur motorcycle racer, averages 129.52 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.29 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 128.32 and 129.65 seconds.

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

d. The middle 50% of her laps are from seconds to seconds

Answer 1

SlutionA:

X~N(129.52 ,2.29 )

SolutionB:

P(128.32<X<129.65)

use belwo R code:

library(tigerstats)
pnormGC(bound=c(128.32,129.65),region="between", mean=129.52 ,sd=2.29,graph=TRUE)

=0.2225

0.2225

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

z score for 4%

is qnorm(0.04) = -1.750686

z= -1.750686

z=x-mean/sd

-1.750686=x-129.52 /2.29

x=-1.750686*2.29+129.52

x= 125.5109

x=125.51

he fastest 4% of laps are under 125.51 seconds

Solutiond:

Q1=0.25

z score is -0.6744898

z=xmean/sd

-0.6744898=x-129.52 /2.29

x=-0.6744898*2.29+129.52

x= 127.9754

For Q3

zscore is

qnorm(0.75)

=0.6744898

z=x-mean/sd

0.6744898=x-129.52 /2.29

x=2.29 *0.6744898+129.52

x= 131.0646

x=131.06

The middle 50% of her laps are from 127.98 seconds to 131.06 seconds.