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At the most recent acceleration survival track meet, Jinwhye\'s obtained the fol

ID: 3125431 • Letter: A

Question

At the most recent acceleration survival track meet, Jinwhye's obtained the following z-scores:

whiplash knot: z = 1.58

dead-end dive: z = -0.74

What was Jinwhye's percentile rank for the dead-end dive? Express your answer to the nearest whole number.

At the most recent acceleration survival track meet, Jinwhye's obtained the following z-scores:

whiplash knot: z = 1.26

dead-end dive: z = -0.41

What was Jinwhye's percentile rank for the dead-end dive? Express your answer to the nearest whole number.

Find the score that is greater than 90% of all other scores in a normal distribution with a mean of 400 and a standard deviation of 50?

A normal distribution has a mean of 400 and a standard deviation of 50, what is the probability (as a percent) of randomly selecting a score between the MEAN and a score of 487?



In a normally-distributed variable measured at the interval/ratio level, approximately what portion of scores will fall between the mean and one standard deviation below the mean?

2/3

1/3

1/2

1/4


To determine the probability of rolling a "2" and a "6" with two dice, we would employ the:

converse rule

proportionality rule

multiplication rule

addition rule

2/3

1/3

1/2

1/4

Explanation / Answer

From the z score table find the corresponding z score for -0.74 which is 0.2296.

Subtract it from 1.00 for the percentile: 1-0.2296=0.7704

Thus, inwhye's percentile rank for the dead-end dive is 77.04 th percentile.

From the z score table find the corresponding z score for -0.41 which is 0.3409.

Subtract it from 1.00 for the percentile: 1-0.3409=0.6591

Thus, inwhye's percentile rank for the dead-end dive is 65.91 th percentile.

In a normally-distributed variable measured at the interval/ratio level, approximately 1/3 rd portion of scores will fall between the mean and one standard deviation below the mean.

To determine the probability of rolling a "2" and a "6" with two dice, we would employ the: addition rule.