91Ó°ÊÓ

In the 2004 baseball season, Ichiro Suzuki of the Seattle Mariners set the record for the most hits in a season with a total of 262 hits. In the following probability distribution, the random variable \(X\) represents the number of hits Ichiro obtained in a game. $$ \begin{array}{ll} x & P(x) \\ \hline 0 & 0.1677 \\ \hline 1 & 0.3354 \\ \hline 2 & 0.2857 \\ \hline 3 & 0.1491 \\ \hline 4 & 0.0373 \\ \hline 5 & 0.0248 \end{array} $$ (a) Verify that this is a discrete probability distribution. (b) Draw a graph of the probability distribution. Describe the shape of the distribution. (c) Compute and interpret the mean of the random variable \(X\). (d) Compute the standard deviation of the random variable \(X\). (e) What is the probability that in a randomly selected game Ichiro got 2 hits? (f) What is the probability that in a randomly selected game Ichiro got more than 1 hit?

Step by step solution

01

Title - Verify Discrete Probability Distribution

To verify that it's a discrete probability distribution, ensure that the sum of all probabilities equals 1 and each probability is between 0 and 1.

02

Calculation to Verify the Distribution

Sum the probabilities: \(0.1677 + 0.3354 + 0.2857 + 0.1491 + 0.0373 + 0.0248 = 1\)Each value is between 0 and 1.

03

Title - Graph the Probability Distribution

Draw a bar graph with the number of hits on the x-axis and their corresponding probabilities on the y-axis.

04

Description of the Shape

The graph is slightly skewed to the right, as the probabilities decrease gradually after 1 hit.

05

Title - Compute the Mean of X

Calculate the mean using the formula for the expected value: \[\text{E}(X) = \sum_x [x \cdot P(x)]\]Multiply each value of x by its probability and sum them up.

06

Calculation of Mean

\(\text{E}(X) = 0 \cdot 0.1677 + 1 \cdot 0.3354 + 2 \cdot 0.2857 + 3 \cdot 0.1491 + 4 \cdot 0.0373 + 5 \cdot 0.0248 = 1.57\)

07

Title - Compute the Standard Deviation of X

Use the formula for standard deviation of a probability distribution: \[\text{SD}(X) = \sqrt{\sum_x [(x - \text{E}(X))^2 \cdot P(x)]}\]

08

Calculation of Standard Deviation

\(\text{SD}(X) = \sqrt{(0 - 1.57)^2 \cdot 0.1677 + (1 - 1.57)^2 \cdot 0.3354 + (2 - 1.57)^2 \cdot 0.2857 + (3 - 1.57)^2 \cdot 0.1491 + (4 - 1.57)^2 \cdot 0.0373 + (5 - 1.57)^2 \cdot 0.0248} \approx 1.12\)

09

Title - Compute P(X=2)

Find the probability that Ichiro got 2 hits in a game. It's given directly in the table: \(P(X=2) = 0.2857\).

10

Title - Compute P(X>1)

Find the probability that Ichiro got more than 1 hit in a game by summing the probabilities of getting 2, 3, 4, and 5 hits.

11

Calculation of P(X>1)

\(P(X>1) = P(X=2) + P(X=3) + P(X=4) + P(X=5) = 0.2857 + 0.1491 + 0.0373 + 0.0248 = 0.4969\)

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Key Concepts

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

Probability Distribution

A probability distribution provides the probabilities of occurrence of different possible outcomes in an experiment. In our specific problem, we're looking at Ichiro Suzuki's number of hits in a game. The random variable, denoted as \(X\), represents these hits. The probabilities for each hit amount are given:

\[\begin{array}{l|l} x & P(x) \ \hline 0 & 0.1677 \ \hline 1 & 0.3354 \ \hline 2 & 0.2857 \ \hline 3 & 0.1491 \ \hline 4 & 0.0373 \ \hline 5 & 0.0248 \end{array}\]
To verify this as a discrete probability distribution, two criteria must be met:

• Each probability must be between 0 and 1.
• The sum of all probabilities must equal 1.

In our table, all probabilities between \(0\) and \(0.3354\) meet the first requirement. Summing these values confirms the second criterion:
\[0.1677 + 0.3354 + 0.2857 + 0.1491 + 0.0373 + 0.0248 = 1\]
Thus, this is a valid discrete probability distribution.

Expected Value

The expected value of a random variable gives a measure of the center of the probability distribution. For discrete distributions, it's calculated by:
\[\text{E}(X) = \sum_x [x \cdot P(x)]\]
In our case, for Ichiro Suzuki's hits, let's multiply each number of hits by its probability and sum them up:

• For \(x = 0\): \(0 \cdot 0.1677 = 0\)
• For \(x = 1\): \(1 \cdot 0.3354 = 0.3354\)
• For \(x = 2\): \(2 \cdot 0.2857 = 0.5714\)
• For \(x = 3\): \(3 \cdot 0.1491 = 0.4473\)
• For \(x = 4\): \(4 \cdot 0.0373 = 0.1492\)
• For \(x = 5\): \(5 \cdot 0.0248 = 0.124\)

Summing these values:
\[0 + 0.3354 + 0.5714 + 0.4473 + 0.1492 + 0.124 = 1.57\]
The expected value (mean) \(\text{E}(X)\) is thus \(1.57\). This means that Ichiro is expected to get approximately \(1.57\) hits per game on average.

Standard Deviation

Standard deviation measures the spread or dispersion of the probability distribution around the mean. For a discrete probability distribution, it's calculated as:
\[\text{SD}(X) = \sqrt{\sum_x [(x - \text{E}(X))^2 \cdot P(x)]}\]
Let's calculate this for Ichiro Suzuki's hits. Let's first find \(\text{E}(X) = 1.57\), and then use it in the standard deviation formula:

• For \(x = 0\): \( (0 - 1.57)^2 \cdot 0.1677 = 0.4135\)
• For \(x = 1\): \( (1 - 1.57)^2 \cdot 0.3354 = 0.1078\)
• For \(x = 2\): \( (2 - 1.57)^2 \cdot 0.2857 = 0.0353\)
• For \(x = 3\): \( (3 - 1.57)^2 \cdot 0.1491 = 0.3241\)
• For \(x = 4\): \( (4 - 1.57)^2 \cdot 0.0373 = 0.2292\)
• For \(x = 5\): \( (5 - 1.57)^2 \cdot 0.0248 = 0.3646\)

Summing these values:
\[0.4135 + 0.1078 + 0.0353 + 0.3241 + 0.2292 + 0.3646 = 1.4745\]
Taking the square root of the result:
\[\sqrt{1.4745} \approx 1.22\]
The standard deviation, \(\text{SD}(X)\), is approximately \(1.22\). This tells us how spread out Ichiro's number of hits can be compared to the mean of \(1.57\) hits per game.