Question 1170687: Mike's Resume shop specializes in creating resumes for students. A recent survey of Mike's shop revealed that a spelling error was made on 11% of the resumes last year. Assuming this rate continues into this month, and that they prepare 305 resumes this month, calculate the following probabilities.
For full marks your answer should be accurate to at least three decimal places.
a) The probability that there are spelling errors on more than 29 resumes.
b) The probability that there are spelling errors on at least 29 resumes.
c) The probability that there are spelling errors on exactly 29 resumes.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem using the binomial distribution and normal approximation.
**Given:**
* Probability of a spelling error (p) = 11% = 0.11
* Number of resumes (n) = 305
**1. Check for Normal Approximation**
* n * p = 305 * 0.11 = 33.55
* n * q = 305 * (1 - 0.11) = 305 * 0.89 = 271.45
Since both np and nq are greater than 10, we can use the normal approximation to the binomial distribution.
**2. Calculate Mean and Standard Deviation**
* Mean (μ) = n * p = 305 * 0.11 = 33.55
* Standard Deviation (σ) = √(n * p * q) = √(305 * 0.11 * 0.89) = √30.0595 ≈ 5.48265
**3. Apply Continuity Correction**
Since we are approximating a discrete distribution (binomial) with a continuous distribution (normal), we need to apply continuity correction.
**(a) Probability of more than 29 resumes with errors (P(X > 29))**
* We need to find P(X > 29.5) using the normal approximation.
1. **Calculate the z-score:**
* z = (x - μ) / σ
* z = (29.5 - 33.55) / 5.48265
* z = -4.05 / 5.48265
* z ≈ -0.7387
2. **Find the probability:**
* We want P(Z > -0.7387) = 1 - P(Z < -0.7387)
* Using a z-table or calculator, P(Z < -0.7387) ≈ 0.2301
* P(Z > -0.7387) = 1 - 0.2301 = 0.7699
3. **Answer:**
* P(X > 29) ≈ 0.770
**(b) Probability of at least 29 resumes with errors (P(X ≥ 29))**
* We need to find P(X > 28.5) using the normal approximation.
1. **Calculate the z-score:**
* z = (x - μ) / σ
* z = (28.5 - 33.55) / 5.48265
* z = -5.05 / 5.48265
* z ≈ -0.9211
2. **Find the probability:**
* We want P(Z > -0.9211) = 1 - P(Z < -0.9211)
* Using a z-table or calculator, P(Z < -0.9211) ≈ 0.1784
* P(Z > -0.9211) = 1 - 0.1784 = 0.8216
3. **Answer:**
* P(X ≥ 29) ≈ 0.822
**(c) Probability of exactly 29 resumes with errors (P(X = 29))**
* We need to find P(28.5 < X < 29.5) using the normal approximation.
1. **Calculate the z-scores:**
* z1 = (28.5 - 33.55) / 5.48265 ≈ -0.9211
* z2 = (29.5 - 33.55) / 5.48265 ≈ -0.7387
2. **Find the probability:**
* P(-0.9211 < Z < -0.7387) = P(Z < -0.7387) - P(Z < -0.9211)
* P(Z < -0.7387) ≈ 0.2301
* P(Z < -0.9211) ≈ 0.1784
* P(-0.9211 < Z < -0.7387) = 0.2301 - 0.1784 = 0.0517
3. **Answer:**
* P(X = 29) ≈ 0.052
**Final Answers:**
a) The probability of more than 29 resumes with errors is approximately 0.770.
b) The probability of at least 29 resumes with errors is approximately 0.822.
c) The probability of exactly 29 resumes with errors is approximately 0.052.
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