91Ó°ÊÓ

Twelve of the top 20 finishers in the 2009 PGA Championship at Hazeltine National Golf Club in Chaska, Minnesota, used a Titleist brand golf ball (GolfBallTest website, November 12,2009 . Suppose these results are representative of the probability that a randomly selected PGA Tour player uses a Titleist brand golf ball. For a sample of 15 PGA Tour players, make the following calculations. a. Compute the probability that exactly 10 of the 15 PGA Tour players use a Titleist brand golf ball. b. Compute the probability that more than 10 of the 15 PGA Tour players use a Titleist brand golf ball. c. For a sample of 15 PGA Tour players, compute the expected number of players who use a Titleist brand golf ball. d. For a sample of 15 PGA Tour players, compute the variance and standard deviation of the number of players who use a Titleist brand golf ball.

Step by step solution

01

Determine the probability of using Titleist

The problem states that 12 out of the top 20 finishers used a Titleist golf ball. Therefore, the probability \( p \) that a randomly selected player uses a Titleist brand golf ball is \( \frac{12}{20} = 0.6 \).

02

Set up binomial distribution parameters

We want to analyze a sample of 15 players, which corresponds to \( n = 15 \), and the probability of success (a player using a Titleist ball) is \( p = 0.6 \). This fits a binomial distribution \( B(n, p) \).

03

Probability of exactly 10 players using Titleist

The probability that exactly 10 out of 15 players use a Titleist ball is found using the binomial probability formula: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). Substituting the values, we get \( P(X = 10) = \binom{15}{10} (0.6)^{10} (0.4)^{5} \). Calculate this to find the probability for part (a).

04

Probability of more than 10 players using Titleist

For more than 10 players, calculate \( P(X > 10) \) using complementary probability: \[ P(X > 10) = 1 - P(X \leq 10) = 1 - (P(X = 0) + P(X = 1) + \ldots + P(X = 10)). \] Calculate each term and subtract their sum from 1 for part (b).

05

Expected number of players using Titleist

For binomial distributions, the expected number \( E(X) \) of players using a Titleist ball is given by \( E(X) = n \times p = 15 \times 0.6 = 9 \). This results in part (c).

06

Variance and standard deviation

The variance \( \sigma^2 \) of a binomial distribution is \( \sigma^2 = n \times p \times (1-p) = 15 \times 0.6 \times 0.4 = 3.6 \). The standard deviation \( \sigma \) is the square root of the variance: \( \sigma = \sqrt{3.6} \approx 1.897 \). This addresses part (d).

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Key Concepts

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

Probability

Probability is a fundamental concept in statistics that measures the likelihood of an event occurring. In this exercise, we are dealing with binomial probability, which is used when there are two possible outcomes.
When considering whether a PGA Tour player uses a Titleist brand golf ball, the "success" is defined as a player using that brand.
Here, the probability, denoted as \( p \), is calculated using the given data: 12 out of 20 players, giving us \( p = \frac{12}{20} = 0.6 \).
This probability helps us understand the likelihood of an event repeating under the same circumstances.

• Key point: Probability \( p = 0.6 \) signifies a 60% chance of any player using a Titleist ball.
• If you were to randomly select one player, there is a 60% likelihood they use the Titleist brand.

Expected Value

The expected value, often denoted as \( E(X) \), is a measure of the center of a probability distribution. It represents the average outcome if one were to conduct the experiment infinitely.
For a binomial distribution, the expected value can be calculated using the formula \( E(X) = n \times p \).
In the given exercise, for 15 PGA Tour players (\( n = 15 \)), the expected number of those using a Titleist ball is \( E(X) = 15 \times 0.6 = 9 \).

• This means that on average, we would expect 9 out of 15 players to use a Titleist brand golf ball.
• The expected value gives us a basis for expectation over the course of many samples.

Variance

Variance is a statistic that measures how far a set of numbers is spread out. For a binomial distribution, variance (denoted \( \sigma^2 \)) helps us understand the variability or dispersion of the data.
It is calculated by the formula \( \sigma^2 = n \times p \times (1-p) \).
Using our example, the variance of players using a Titleist ball is \( \sigma^2 = 15 \times 0.6 \times 0.4 = 3.6 \).

• The variance tells us that there is some degree of fluctuation around the expected number of players, which is the natural variability in any sample.
• A high variance would indicate more spread out results, while a low variance would suggest results are closely clustered.

Standard Deviation

Standard deviation is a measure of how much we can expect the data to deviate from the expected value, representing variability in the same units as the data.
It is the square root of the variance: \( \sigma = \sqrt{\sigma^2} \).
In this case, the standard deviation is \( \sigma = \sqrt{3.6} \approx 1.897 \).

• A standard deviation of roughly 1.897 means that in the context of this experiment, the number of players using a Titleist ball typically varies by about 1.897 players around the expected value (9).
• Understanding standard deviation helps us grasp the normal range of variability from the average.