91Ó°ÊÓ

In college basketball, a shot made from beyond a designated arc radiating about 20 feet from the basket is worth three points instead of the usual two points given for shots made inside that arc. Over his career, University of Florida basketball player Lee Humphrey made \(45 \%\) of his three-point attempts. In one game in his final season, he made only 3 of 12 three-point shots, leading a TV basketball analyst to announce that Humphrey was in a shooting slump. a. Assuming Humphrey has a \(45 \%\) chance of making any particular three-point shot, find the mean and standard deviation of the sampling distribution of the proportion of three-point shots he will make out of 12 shots. b. How many standard deviations from the mean is this game's result of making 3 of 12 three-point shots? c. If Humphrey was actually not in a slump but still had a \(45 \%\) chance of making any particular three-point shot, explain why it would not be especially surprising for him to make only 3 of 12 shots. Thus, this is not really evidence of a shooting slump.

Step by step solution

01

Determine the mean of the distribution

The mean of the distribution, denoted by \( \mu \), can be found using the formula \( \mu = np \), where \( n \) is the number of shots and \( p \) is the probability of making a shot. Here, \( n = 12 \) and \( p = 0.45 \). Thus, \( \mu = 12 \times 0.45 = 5.4 \).

02

Calculate the standard deviation of the distribution

The standard deviation, denoted by \( \sigma \), can be calculated using the formula \( \sigma = \sqrt{np(1-p)} \). Substituting the values, \( \sigma = \sqrt{12 \times 0.45 \times 0.55} \). This simplifies to \( \sigma = \sqrt{2.97} \approx 1.723 \).

03

Calculate the z-score for making 3 out of 12 shots

The z-score is calculated using the formula \( z = \frac{X - \mu}{\sigma} \), where \( X \) is the observed value for the number of successful shots. Here, \( X = 3 \). Substituting the known values, \( z = \frac{3 - 5.4}{1.723} \approx -1.393 \).

04

Interpret the z-score and assess the likelihood

A z-score of approximately \(-1.393\) means that making 3 out of 12 shots is \(1.393\) standard deviations below the mean. In a normal distribution, a z-score between \(-2\) and \(2\) is not unusual, indicating that this outcome is within a typical range and not surprising.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Calculation

Calculating the mean is a fundamental step in statistical analysis. It helps us find the expected or average outcome. In the context of Lee Humphrey's basketball performance, the mean represents the average number of successful three-point shots he might expect to make out of 12 attempts.

To find the mean, we use the formula:

• \( \mu = np \)

Here, \( n \) represents the number of attempts (12 shots), and \( p \) is the probability of making a shot (45% or 0.45). Plugging in these values gives us:

• \( \mu = 12 \times 0.45 = 5.4 \)

Thus, Humphrey could expect to make about 5.4 shots on average, which means he generally hits between 5 to 6 shots given the number of attempts.

Standard Deviation

Standard deviation is a measure of how spread out numbers are around the mean. In our basketball case, it tells us how much Humphrey's actual number of successful shots might vary from the average.

The formula to calculate standard deviation in this context is:

• \( \sigma = \sqrt{np(1-p)} \)

Again, using \( n = 12 \) and \( p = 0.45 \), we calculate:

• \( \sigma = \sqrt{12 \times 0.45 \times 0.55} \approx 1.723 \)

A standard deviation of approximately 1.723 indicates how much we can expect his performance to "normally" vary on either side of the mean of 5.4 shots in any given set of 12 attempts.

Z-score

The z-score helps us determine how unusual or typical a particular value is within a data set. It tells us how many standard deviations away an observation is from the mean.

For Humphrey's result where he makes 3 successful shots out of 12, we can calculate the z-score as follows:

• \( z = \frac{X - \mu}{\sigma} \)

Here, \( X = 3 \) (actual shots made), \( \mu = 5.4 \) (the mean), and \( \sigma = 1.723 \) (standard deviation). Substituting these into the formula gives:

• \( z = \frac{3 - 5.4}{1.723} \approx -1.393 \)

A z-score of approximately \(-1.393\) reveals that making 3 shots is 1.393 standard deviations below the mean. In a normal distribution context, results within two standard deviations are common, so this outcome is not unusual.

Sampling Distribution

A sampling distribution shows the behavior of sample statistics over multiple samples from the same population. In this case, it represents the distribution of the proportion of successful shots Humphrey makes out of a series of games, each consisting of 12 attempts.

If Humphrey’s success rate over time is 45%, sampling distribution can help us understand his performance. For any such game, by examining where the mean (5.4 shots) and standard deviation (1.723 shots) fit, we can predict the range of likely outcomes.

Even if Humphrey hits only 3 shots in one game, given the normal spread of outcomes, it's not necessarily a slump. Results like this fall within typical variation that the sampling distribution anticipates if his long-run success rate holds at 45%. With this context, one can see that subpar or stellar performances over a short run don't always imply lasting trends.