Answer :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 129.01, \sigma = 2.26[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{130 - 129.01}{2.26}[/tex]
[tex]Z = 0.44[/tex]
[tex]Z = 0.44[/tex] 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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.88 = \frac{X - 129.01}{2.26}[/tex]
[tex]X - 129.01 = -1.88*2.26[/tex]
[tex]X = 124.8[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.28 = \frac{X - 129.01}{2.26}[/tex]
[tex]X - 129.01 = -1.28*2.26[/tex]
[tex]X = 126.1[/tex]
90th percentile:
X when Z has a pvalue of 0.9, so X when Z = 1.28.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 129.01}{2.26}[/tex]
[tex]X - 129.01 = 1.28*2.26[/tex]
[tex]X = 131.9[/tex]
The middle 80% of her laps are from 126.1 seconds to 131.9 seconds.