Question

Terri Vogel, an amateur motorcycle racer, averages 129.01 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.26 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.
Required:
a. In words, define the random variable X.
b. X ~ ________(_____ , _____)
c. Find the percent of her laps that are completed in less than 130 seconds.
d. The fastest 3% of her laps are under
e. The middle 80% of her laps are from ___________seconds to ________seconds.

Answer 1

Answer:

a. X is the time, in seconds, of a randomly selected lap of Vogel.

b. X~N(129.01,2.26)

c. 67% of her laps that are completed in less than 130 seconds.

d. So the answer is 124.8 seconds.

e. The middle 80% of her laps are from 126.1 seconds to 131.9 seconds.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Terri Vogel, an amateur motorcycle racer, averages 129.01 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.26 seconds

This means that $\mu = 129.01, \sigma = 2.26$

a. In words, define the random variable X.

The problem states that X is the time, in seconds, of a randomly selected lap of Vogel.

b. X ~ ________(_____ , _____)

Normal with mean and standard deviation. So

X~N(129.01,2.26)

c. Find the percent of her laps that are completed in less than 130 seconds.

The proportion is the pvalue of Z when X = 130. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{130 - 129.01}{2.26}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67.

0.67*100% = 67%

67% of her laps that are completed in less than 130 seconds.

d. The fastest 3% of her laps are under

Under the 3rd percentile, which is X when Z has a pvalue of 0.03. So X when Z = -1.88.

$Z = \frac{X - \mu}{\sigma}$

$-1.88 = \frac{X - 129.01}{2.26}$

$X - 129.01 = -1.88*2.26$

$X = 124.8$

So the answer is 124.8 seconds.

e. The middle 80% of her laps are from ___________seconds to ________seconds.

The 50 - (80/2) = 10th percentile to the 50 + (80/2) = 90th percentile.

10th percentile:

X when Z has a pvalue of 0.1, so X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 129.01}{2.26}$

$X - 129.01 = -1.28*2.26$

$X = 126.1$

90th percentile:

X when Z has a pvalue of 0.9, so X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 129.01}{2.26}$

$X - 129.01 = 1.28*2.26$

$X = 131.9$

The middle 80% of her laps are from 126.1 seconds to 131.9 seconds.