Math Problem Statement

Independent record stores, while declining in number, had not completely vanished in the early 2020s. Your company, Video Records, Inc., has conducted research that shows the following probability distribution, where X is the number of record stores in a randomly chosen city with more than 100,000 inhabitants. x 1 2 3 4 5 6 7 8 9 10 P(X = x) 0.01 0.01 0.01 0.02 0.08 0.14 0.27 0.27 0.14 0.05 (a) Compute πœ‡ = E(X). HINT [See Example 3.] E(X) = Interpret the result. There are at most this many record stores in each city with more than 100,000 inhabitants. There are, on average, this many record stores in a city with more than 100,000 inhabitants.
This is the most frequently observed number of record stores in cities with more than 100,000 inhabitants. There are at least this many record stores in each city with more than 100,000 inhabitants. (b) Find P(X < πœ‡) or P(X > πœ‡). P(x < πœ‡)

P(x > πœ‡)

Interpret the result. The P(x > πœ‡) is ---Select---

P(x < πœ‡). Thus, most cities have ---Select--- the average number of record stores.

Solution

(a) Compute ΞΌ=E(X)\mu = E(X)

The expected value (mean) E(X)E(X) is computed using the formula:

E(X)=βˆ‘xxβ‹…P(X=x)E(X) = \sum_{x} x \cdot P(X = x)

Given the values from the probability distribution:

\hline x & P(X = x) \\ \hline 1 & 0.01 \\ 2 & 0.01 \\ 3 & 0.01 \\ 4 & 0.02 \\ 5 & 0.08 \\ 6 & 0.14 \\ 7 & 0.27 \\ 8 & 0.27 \\ 9 & 0.14 \\ 10 & 0.05 \\ \hline \end{array}$$ Now, calculate $$E(X)$$: $$E(X) = (1 \cdot 0.01) + (2 \cdot 0.01) + (3 \cdot 0.01) + (4 \cdot 0.02) + (5 \cdot 0.08) + (6 \cdot 0.14) + (7 \cdot 0.27) + (8 \cdot 0.27) + (9 \cdot 0.14) + (10 \cdot 0.05)$$ $$E(X) = 0.01 + 0.02 + 0.03 + 0.08 + 0.40 + 0.84 + 1.89 + 2.16 + 1.26 + 0.50$$ $$E(X) = 7.23$$ Thus, the expected number of record stores in a randomly chosen city with more than 100,000 inhabitants is **7.23**. ### Interpretation of the result: - The result means that, **on average**, there are approximately **7.23 record stores** in each city with more than 100,000 inhabitants. - This is not necessarily the most frequent number of stores, but an average across all possible values of $$X$$. ### (b) Find $$P(X < \mu)$$ or $$P(X > \mu)$$ Now, we need to find the probabilities of $$X$$ being less than 7.23 and greater than 7.23: 1. **$$P(X < 7.23)$$:** This is the probability that the number of record stores is less than the expected value. $$P(X < 7.23) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$$ $$P(X < 7.23) = 0.01 + 0.01 + 0.01 + 0.02 + 0.08 + 0.14 = 0.27$$ 2. **$$P(X > 7.23)$$:** This is the probability that the number of record stores is greater than the expected value. $$P(X > 7.23) = P(X = 8) + P(X = 9) + P(X = 10)$$ $$P(X > 7.23) = 0.27 + 0.14 + 0.05 = 0.46$$ ### Interpretation of the result: - $$P(X < 7.23) = 0.27$$ means that **27%** of the cities have fewer than 7.23 record stores. - $$P(X > 7.23) = 0.46$$ means that **46%** of the cities have more than 7.23 record stores. - Thus, **more cities** have more than the average number of record stores than have fewer. --- ### Summary: - $$\mu = E(X) = 7.23$$ - $$P(X < 7.23) = 0.27$$ - $$P(X > 7.23) = 0.46$$ ### Next questions: 1. How would the expected value change if the probability distribution shifted towards more record stores in each city? 2. How would you calculate the variance of this probability distribution? 3. What would the probability distribution look like if the number of record stores were uniformly distributed? 4. How do you interpret the difference between $$P(X < 7.23)$$ and $$P(X > 7.23)$$? 5. What other metrics (like mode or median) might provide additional insights into the distribution? ### Tip: In probability, understanding the difference between the **mean** (average) and **mode** (most frequent value) can give you deeper insight into how data is distributed. In your case, while the average number of record stores is 7.23, this does not tell you the most common number of stores; it simply represents the center of the distribution.

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Math Problem Analysis

Mathematical Concepts

Probability Distributions
Expected Value
Basic Statistics

Formulas

E(X) = Ξ£ x * P(X = x)

Theorems

Expected Value Theorem

Suitable Grade Level

Grades 10-12

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