A computer store knows from past experience that 11% of computers bought will be returned for a full refund. A random sample of 31 computer purchases is selected at random. Assume that the binomial distribution applies. Note: The mean and standard deviation of a binomial random variable are given by A. What is the number of trials? Answer: B. What is the probability of "s "success"? C 4 $ Answer: C. What is the mean number of computer purchases out of 31 that will be returned for a refund? Round to 1 decimal place. D. What is the standard deviation of the number of computer purchases out of 31 that will be returned for a refunds Answer: % 5 ^ 6 Oll Do not round. μ = np, no & 7 O * o = √npq 8 O ( 9 41 ) 0 → 4 11 E

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### Binomial Distribution in Computer Purchases

A computer store knows from past experience that 11% of computers bought will be returned for a full refund. A random sample of 31 computer purchases is selected at random. Assume that the binomial distribution applies.

**Note:** The mean and standard deviation of a binomial random variable are given by:

\[ \mu = np \]
\[ \sigma = \sqrt{npq} \]

Where:
- \( n \) is the number of trials,
- \( p \) is the probability of success,
- \( q = 1 - p \) is the probability of failure.

#### Questions:

**A. What is the number of trials?**  
**Answer:**  
\[ \_\_\_\_\_\_\_\_\_ \]

**B. What is the probability of "success"?**  
**Answer:**  
*Do not round.*

\[ \_\_\_\_\_\_\_\_\_ \]

**C. What is the mean number of computer purchases out of 31 that will be returned for a refund?**  
**Answer:**  
*Round to 1 decimal place.*

\[ \_\_\_\_\_\_\_\_\_ \]

**D. What is the standard deviation of the number of computer purchases out of 31 that will be returned for a refund?**  
**Answer:**  
\[ \_\_\_\_\_\_\_\_\_ \]

---

**Explanation of Formulae:**

- **Mean (μ):** The average number of successes in n trials, calculated as \( \mu = np \). For this problem, it represents the expected number of computer returns.
- **Standard Deviation (σ):** Measures the spread of the distribution, calculated as \( \sigma = \sqrt{npq} \).
Transcribed Image Text:--- ### Binomial Distribution in Computer Purchases A computer store knows from past experience that 11% of computers bought will be returned for a full refund. A random sample of 31 computer purchases is selected at random. Assume that the binomial distribution applies. **Note:** The mean and standard deviation of a binomial random variable are given by: \[ \mu = np \] \[ \sigma = \sqrt{npq} \] Where: - \( n \) is the number of trials, - \( p \) is the probability of success, - \( q = 1 - p \) is the probability of failure. #### Questions: **A. What is the number of trials?** **Answer:** \[ \_\_\_\_\_\_\_\_\_ \] **B. What is the probability of "success"?** **Answer:** *Do not round.* \[ \_\_\_\_\_\_\_\_\_ \] **C. What is the mean number of computer purchases out of 31 that will be returned for a refund?** **Answer:** *Round to 1 decimal place.* \[ \_\_\_\_\_\_\_\_\_ \] **D. What is the standard deviation of the number of computer purchases out of 31 that will be returned for a refund?** **Answer:** \[ \_\_\_\_\_\_\_\_\_ \] --- **Explanation of Formulae:** - **Mean (μ):** The average number of successes in n trials, calculated as \( \mu = np \). For this problem, it represents the expected number of computer returns. - **Standard Deviation (σ):** Measures the spread of the distribution, calculated as \( \sigma = \sqrt{npq} \).
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