Question

paper. be as stupid specific as possible. we will grade this in class, and yes, it will count as a grade. do not write your name on your paper. please write your birth date or student id. you may use your notes and work with your partner. each student needs to turn this in.
a professional sports team evaluates potential players for a certain position based on two main characteristics, speed and strength.
(a) speed is measured by the time required to run a distance of 40 yards, with smaller times indicating more desirable (faster) speeds. from previous speed data for all players in this position, the times to run 40 yards have a mean of 4.60 seconds and a standard deviation of 0.15 seconds, with a minimum time of 4.40 seconds, as shown in the table below.

based on the relationship between the mean, standard deviation, and minimum time, is it reasonable to believe that the distribution of 40 - yard running times is approximately normal? explain.
(b) strength is measured by the amount of weight lifted, with more weight indicating more desirable (greater) strength. from previous strength data for all players in this position, the amount of weight lifted has a mean of 310 pounds and a standard deviation of 25 pounds, as shown in the table below.

calculate and interpret the z - score for a player in this position who can lift a weight of 370 pounds.
(c) the characteristics of speed and strength are considered to be of equal importance to the team in selecting a player for the position. based on the information about the means and standard deviations of the speed and strength data for all players and the measurements listed in the table below for players a and b, which player should the team select if the team can only select one of the two players? justify your answer.

Explanation:

Part (a)

Step1: Recall the properties of normal distribution

In a normal distribution, about \(99.7\%\) of the data lies within \(3\) standard deviations of the mean. The lower bound for \(3\) standard deviations below the mean is \(\mu - 3\sigma\), where \(\mu = 4.60\) and \(\sigma=0.15\).

Step2: Calculate \(\mu - 3\sigma\)

$$ \mu - 3\sigma=4.60-3\times0.15 = 4.60 - 0.45=4.15 $$

The minimum time given is \(4.40\) seconds. Since \(4.15<4.40\), the distribution is likely skewed (because the lower - tail is “cut off” at \(4.40\) instead of extending to \(4.15\) as would be expected in a normal distribution).

Step1: Recall the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the observed value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step2: Substitute the values into the formula

Given \(x = 370\), \(\mu=310\), and \(\sigma = 25\)

$$ z=\frac{370 - 310}{25}=\frac{60}{25}=2.4 $$

Step3: Interpret the z - score

A z - score of \(z = 2.4\) means that the weight lifted by the player is \(2.4\) standard deviations above the mean weight lifted by all players in that position.

Step1: Calculate the z - score for speed (Player A)

For speed (smaller time is better), \(\mu = 4.60\), \(\sigma=0.15\), \(x = 4.42\)

$$ z_{A - speed}=\frac{4.42-4.60}{0.15}=\frac{- 0.18}{0.15}=-1.2 $$

For Player B, \(x = 4.57\)

$$ z_{B - speed}=\frac{4.57 - 4.60}{0.15}=\frac{-0.03}{0.15}=- 0.2 $$

Step2: Calculate the z - score for strength (Player A)

For strength (larger weight is better), \(\mu = 310\), \(\sigma = 25\), \(x = 370\) (already calculated \(z_{A - strength}=2.4\) in part (b))
For Player B, \(x = 375\)

$$ z_{B - strength}=\frac{375-310}{25}=\frac{65}{25}=2.6 $$

Step3: Calculate the combined z - score (assuming equal weight for speed and strength)

For Player A: \(z_{A}=\frac{z_{A - speed}+z_{A - strength}}{2}=\frac{-1.2 + 2.4}{2}=0.6\)
For Player B: \(z_{B}=\frac{z_{B - speed}+z_{B - strength}}{2}=\frac{-0.2+2.6}{2}=1.2\)

Answer:

It is not reasonable to believe that the distribution of 40 - yard running times is approximately normal. Because for a normal distribution, we would expect about \(99.7\%\) of the data to be within \(3\) standard deviations of the mean (\(\mu - 3\sigma=4.15\) seconds), but the minimum time is \(4.40\) seconds.

Part (b)