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Chapter 5: Problem 108

A certain professional basketball player typically makes \(80 \%\) of his basket attempts, which is considered to be good. Suppose you go to several games at which this player plays. Sometimes the player attempts only a few baskets, say, 10. Other times, he attempts about 60 . On which of those nights is the player most likely to have a "bad" night, in which he makes much fewer than \(80 \%\) of his baskets?

Short Answer

Expert verified
The player is more likely to have a 'bad' night when he makes around 60 attempts compared to when he only tries 10 because the variability or standard deviation is greater in the former case.

Step by step solution

01

Determine the Mean

First, calculate the expected number of baskets (mean) the player will typically make. For 10 attempts, the expected number of made shots is $0.8 \times 10 = 8$. Similarly, for 60 attempts, it is $0.8 \times 60 = 48$.
02

Calculate the Variance and Standard deviation

The variance of a Bernoulli distribution is given by $np(1-p)$. So, for 10 attempts, the variance is $10 \times 0.8 \times (1 - 0.8) = 1.6 $. For 60 attempts, it is $60 \times 0.8 \times (1 - 0.8) = 9.6 $. The standard deviation, which shows the extent of variability, is the square root of the variance. So, for 10 attempts, the standard deviation is $\sqrt{1.6} \approx 1.27 $. For 60 attempts, it is $\sqrt{9.6} \approx 3.1 $.
03

Compare and Conclude

From the calculation above, we notice that the standard deviation for 60 attempts is higher. This means there's a higher variability, in other words, a greater chance of making much fewer than \(80 \% \) of the shots, i.e., having a 'bad' night.

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Key Concepts

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

Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the values in a data set deviate from the mean or expected value. The formula for the standard deviation is the square root of the variance. In simpler terms,
  • It helps us understand if the values (number of baskets made, in this case) are close to the average or spread out over a wide range.
  • A smaller standard deviation means that the values tend to be close to the mean, indicating less variability.
  • A larger standard deviation indicates more spread out values, showing more variability.

In our basketball example, the standard deviation is higher with 60 attempts compared to 10 attempts. This suggests that for more shots, the outcomes are spread out more broadly around the mean of 48 successful shots. This increased variability means that there is a greater chance of making significantly fewer baskets than expected.
Variance
Variance is closely related to the standard deviation, as it is essentially the standard deviation squared. It measures how far the data points (like the number of made baskets) are spread out from their average value. To calculate the variance in a Bernoulli distribution, which is applicable here since only two outcomes are possible (making or missing a basket), use the formula \[ np(1-p) \] where:
  • \( n \) is the number of trials (or attempts).
  • \( p \) is the probability of success (basket made, so 0.8 here).

For 10 attempts, the variance is 1.6, and for 60 attempts, it is 9.6.
This means the broader range of values (e.g., how many baskets are made) is covered when the player takes 60 shots, leading to a higher potential for variability.
Expected Value
The expected value is a fundamental concept in probability that gives us the average outcome of a random event if the experiment were repeated many times. In the context of basketball shots, it tells us how many shots the player is expected to make on average.
  • The expected value can be found using the formula \( np \), where \( n \) is the number of trials (or attempts) and \( p \) is the probability of success (making a basket, 0.8 in this case).

So, when the player attempts 10 shots, the expected number of made baskets is \( 0.8 \times 10 = 8 \). For 60 attempts, it is \( 0.8 \times 60 = 48 \). This expected value helps to set a baseline, representing the average performance we anticipate based on probability, and it's crucial for assessing how much actual performance deviates, which is exactly what variance and standard deviation measure.

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