# Fatigue Factor of Safety Calculation (Modified Goodman Criterion) ## Given Information - **Material:** Cold-drawn AISI 1018 steel - **Shaft Diameter:** \( d = 20 \) mm - **Critical Stress Element:** - Bending stress (\( \sigma \)) = 77.5 MPa - Torsional stress (\( \tau \)) = 20 MPa - **Shaft Rotation:** Constant speed --- ## Objective **Determine the minimum factor of safety for fatigue (infinite life) using the Modified Goodman criterion.** --- ## Key Concepts - **Bending stress** is fully reversed (alternating). - **Torsional stress** is steady (mean). - **Combined stresses:** Use von Mises theory for equivalent alternating and mean stress: - \( \sigma'_a = \sqrt{\sigma_a^2 + 3\tau_a^2} \) - \( \sigma'_m = \sqrt{\sigma_m^2 + 3\tau_m^2} \) - **Modified Goodman Criterion:** \[ \frac{\sigma'_a}{S_e} + \frac{\sigma'_m}{S_u} = \frac{1}{n} \] where: \( S_e \) = endurance limit (corrected), \( S_u \) = ultimate strength, \( n \) = factor of safety --- ## Step 1: Identify Alternating and Mean Components - **Bending:** - Alternating: \( \sigma_a = 77.5 \) MPa - Mean: \( \sigma_m = \) MPa - **Torsion:** - Alternating: \( \tau_a = \) MPa - Mean: \( \tau_m = 20 \) MPa --- ## Step 2: Equivalent Stresses - **Equivalent alternating stress:** \[ \sigma'_a = \sqrt{(77.5)^2 + 3 \times ()^2} = 77.5 \text{ MPa} \] - **Equivalent mean stress:** \[ \sigma'_m = \sqrt{()^2 + 3 \times (20)^2} = \sqrt{120} = 34.641 \text{ MPa} \] --- ## Step 3: Material Properties - **Ultimate strength:** \( S_u = 440 \) MPa - **Unmodified endurance limit:** \( S'_e = .5 S_u = 220 \) MPa --- ## Step 4: Marin Factors & Corrected Endurance Limit - **Surface factor (\( k_a \)), cold-drawn:** \[ k_a = 4.51 S_u^{-.265} = 4.51 \times (440)^{-.265} = .897 \] - **Size factor (\( k_b \)), bending, \( d = 20 \) mm:** \[ k_b = (d/7.62)^{-.107} = (20/7.62)^{-.107} = .902 \] - **Other Marin factors:** ≈ 1 - **Corrected endurance limit:** \[ S_e = k_a k_b S'_e = (.897)(.902)(220) = 177.99 \text{ MPa} \] --- ## Step 5: Apply Modified Goodman Criterion \[ \frac{\sigma'_a}{S_e} + \frac{\sigma'_m}{S_u} = \frac{77.5}{177.99} + \frac{34.641}{440} \] \[ = .4355 + .0787 = .5142 \] - **Factor of safety:** \[ n = \frac{1}{.5142} = 1.9447 \] --- ## **Final Answer** ### The minimum factor of safety for fatigue, based on infinite life using the Modified Goodman criterion, is: \[ \boxed{1.94} \] (Rounded to two decimal places)

# Fatigue Factor of Safety Calculation (Modified Goodman Criterion) ## Given Information - **Material:** Cold-drawn AISI 1018 steel - **Shaft Diameter:** \( d = 20 \) mm - **Critical Stress Element:** - Bending stress (\( \sigma \)) = 77.5 MPa - Torsional stress (\( \tau \)) = 20 MPa - **Shaft Rotation:** Constant speed --- ## Objective **Determine the minimum factor of safety for fatigue (infinite life) using the Modified Goodman criterion.** --- ## Key Concepts - **Bending stress:** Fully reversed (alternating). - **Torsional stress:** Steady (mean). - **Combined stresses:** Use von Mises theory for equivalent alternating and mean stress: - \( \sigma'_a = \sqrt{\sigma_a^2 + 3\tau_a^2} \) - \( \sigma'_m = \sqrt{\sigma_m^2 + 3\tau_m^2} \) - **Modified Goodman Criterion:** \[ \frac{\sigma'_a}{S_e} + \frac{\sigma'_m}{S_u} = \frac{1}{n} \] --- ## Step 1: Identify Alternating and Mean Components - **Bending:** - Alternating: \( \sigma_a = 77.5 \) MPa - Mean: \( \sigma_m = 0 \) MPa - **Torsion:** - Alternating: \( \tau_a = 0 \) MPa - Mean: \( \tau_m = 20 \) MPa --- ## Step 2: Equivalent Stresses - **Equivalent alternating stress:** \[ \sigma'_a = \sqrt{(77.5)^2 + 3 \times (0)^2} = 77.5 \text{ MPa} \] - **Equivalent mean stress:** \[ \sigma'_m = \sqrt{(0)^2 + 3 \times (20)^2} = \sqrt{1200} = 34.641 \text{ MPa} \] --- ## Step 3: Material Properties - **Ultimate strength:** \( S_u = 440 \) MPa - **Unmodified endurance limit:** \( S'_e = 0.5 S_u = 220 \) MPa --- ## Step 4: Marin Factors & Corrected Endurance Limit - **Surface factor (\( k_a \)), cold-drawn:** \[ k_a = 4.51 S_u^{-0.265} = 4.51 \times (440)^{-0.265} = 0.897 \] - **Size factor (\( k_b \)), bending, \( d = 20 \) mm:** \[ k_b = \left( \frac{d}{7.62} \right)^{-0.107} = \left( \frac{20}{7.62} \right)^{-0.107} = 0.902 \] - **Other Marin factors:** ≈ 1 - **Corrected endurance limit:** \[ S_e = k_a k_b S'_e = (0.897)(0.902)(220) = 177.99 \text{ MPa} \] --- ## Step 5: Apply Modified Goodman Criterion \[ \frac{\sigma'_a}{S_e} + \frac{\sigma'_m}{S_u} = \frac{77.5}{177.99} + \frac{34.641}{440} \] \[ = 0.4355 + 0.0787 = 0.5142 \] - **Factor of safety:** \[ n = \frac{1}{0.5142} = 1.9447 \] --- ## Final Answer ### The minimum factor of safety for fatigue, based on infinite life using the Modified Goodman criterion, is: \[ \boxed{1.94} \] (Rounded to two decimal places)

# Weight Consistency Analysis for Packaged Rice Bags ## Given Information - **Distribution of bag weight:** Normal - **Mean (\( \mu \)):** 1000 grams - **Standard deviation (\( \sigma \)):** 12 grams - **Regulatory guideline:** No more than 2% of bags should weigh less than 975 grams - **Overfill threshold:** Bags exceeding 1025 grams are considered inefficient - **Sample size:** \( n = 36 \) --- ## Objectives 1. Determine the probability that a randomly selected bag weighs less than 975 grams. 2. Determine the probability that a randomly selected bag weighs more than 1025 grams. 3. Compute the probability that the average weight of the 36 selected bags is less than 995 grams. --- ## Step 1: Probability of Weighing Less Than 975 grams ### Standardization To find \( P(X < 975) \): - Calculate the z-score: \[ z = \frac{x - \mu}{\sigma} = \frac{975 - 1000}{12} = \frac{-25}{12} \approx -2.0833 \] ### Probability Calculation Using the standard normal distribution table or calculator: \[ P(Z < -2.0833) \approx 0.0187 \] --- ## Step 2: Probability of Weighing More Than 1025 grams ### Standardization To find \( P(X > 1025) \): - Calculate the z-score: \[ z = \frac{1025 - 1000}{12} = \frac{25}{12} \approx 2.0833 \] ### Probability Calculation Using the standard normal distribution: \[ P(Z > 2.0833) = 1 - P(Z < 2.0833) \approx 1 - 0.9813 = 0.0187 \] --- ## Step 3: Probability that the Sample Mean is Less Than 995 grams ### Standardization The standard deviation of the sample mean (\( \sigma_X \)) is: \[ \sigma_X = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2 \] To find \( P(\bar{X} < 995) \): - Calculate the z-score: \[ z = \frac{995 - 1000}{2} = \frac{-5}{2} = -2.5 \] ### Probability Calculation Using the standard normal distribution: \[ P(Z < -2.5) \approx 0.0062 \] --- ## Final Answers 1. \( P(X < 975) \approx 0.0187 \) 2. \( P(X > 1025) \approx 0.0187 \) 3. \( P(\bar{X} < 995) \approx 0.0062 \) All probabilities are rounded to four decimal places.

# Monthly Claim Amount Analysis for Insurance Policies ## Given Information - **Mean monthly claim amount (\( \mu \))**: ₹12,000 - **Standard deviation (\( \sigma \))**: ₹3,000 - **Threshold for "unusually high" claims**: ₹18,000 - **Claim range for reasonable fluctuation**: ₹9,000 to ₹15,000 --- ## Objectives 1. Determine the probability that a randomly selected policyholder's monthly claim amount exceeds ₹18,000. 2. Determine the probability that the monthly claim amount lies between ₹9,000 and ₹15,000. --- ## Concept Used Assuming the distribution of monthly claims (\( Z \)) is approximately normal, we will standardize the values using the z-score formula: \[ z = \frac{x - \mu}{\sigma} \] Then, we will use the standard normal distribution to find the probabilities associated with the standardized z-scores. --- To find \( P(Z > 18000) \): - **Standardization** \[ z_{high} = \frac{18000 - 12000}{3000} = \frac{6000}{3000} = 2 \] - **Probability Calculation** \[ P(Z > 18000) = 1 - P(Z < 2) \] From standard normal distribution tables: \[ P(Z < 2) \approx 0.9772 \] Therefore: \[ P(Z > 18000) \approx 1 - 0.9772 = 0.0228 \] --- To find \( P(9000 < Z < 15000) \): - **Standardization** \[ z_{low} = \frac{9000 - 12000}{3000} = \frac{-3000}{3000} = -1 \] \[ z_{high} = \frac{15000 - 12000}{3000} = \frac{3000}{3000} = 1 \] - **Probability Calculation** \[ P(9000 < Z < 15000) = P(Z < 1) - P(Z < -1) \] From standard normal distribution tables: \[ P(Z < 1) \approx 0.8413 \] \[ P(Z < -1) \approx 0.1587 \] Therefore: \[ P(9000 < Z < 15000) \approx 0.8413 - 0.1587 = 0.6826 \] --- ## Final Answers 1. \( P(Z > 18000) \approx 0.0228 \) 2. \( P(9000 < Z < 15000) \approx 0.6826 \) All probabilities are rounded to four decimal places.

# Reliability Evaluation of Satellite-Based Environmental Monitoring System ## Given Information - **Probability of successful transmission (\( p \))**: 0.97 - **Probability of failure (\( q \))**: \( 1 - p = 0.03 \) - **Monitoring period**: 24 hours - **Satisfactory day** condition: At least 22 successful transmissions - **Total attempts in 24 hours**: \( n = 24 \) --- ## Objectives 1. Determine the probability that a randomly selected 24-hour period is satisfactory. 2. Determine the probability that exactly 3 transmissions fail during the 24-hour period. 3. Determine the probability that at least one day is not satisfactory during a 7-day evaluation. --- ## Concept Used The number of successful transmissions \( X \) follows a **binomial distribution**: \[ X \sim \text{Binomial}(n, p) \] where: - \( n = 24 \) (number of trials) - \( p = 0.97 \) (success probability) The probability mass function (PMF) for a binomial distribution is given by: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \] where \( k \) is the number of successes. --- To find \( P(X \geq 22) \): - Calculate \( P(X = 22) \), \( P(X = 23) \), and \( P(X = 24) \): \[ P(X = 22) = \binom{24}{22} (0.97)^{22} (0.03)^{2} \] \[ P(X = 23) = \binom{24}{23} (0.97)^{23} (0.03)^{1} \] \[ P(X = 24) = \binom{24}{24} (0.97)^{24} (0.03)^{0} \] Calculating each term: - For \( P(X = 22) \): \[ P(X = 22) = \binom{24}{22} (0.97)^{22} (0.03)^{2} = 276 \cdot (0.97)^{22} \cdot (0.03)^{2} \approx 276 \cdot 0.5264 \cdot 0.0009 \approx 0.1324 \] - For \( P(X = 23) \): \[ P(X = 23) = \binom{24}{23} (0.97)^{23} (0.03)^{1} = 24 \cdot (0.97)^{23} \cdot (0.03) \approx 24 \cdot 0.5122 \cdot 0.03 \approx 0.0367 \] - For \( P(X = 24) \): \[ P(X = 24) = \binom{24}{24} (0.97)^{24} \cdot (0.03)^{0} = 1 \cdot (0.97)^{24} \approx 0.4870 \] Now, add these probabilities for satisfactory days: \[ P(X \geq 22) = P(X = 22) + P(X = 23) + P(X = 24) \approx 0.1324 + 0.0367 + 0.4870 \approx 0.6561 \] --- To find \( P(\text{exactly 3 failures}) \): Let \( Y \) denote the number of failures: \[ Y \sim \text{Binomial}(n, q) \] where \( q = 0.03 \). \[ P(Y = 3) = \binom{24}{3} (0.03)^3 (0.97)^{21} \] Calculating this: \[ P(Y = 3) = \binom{24}{3} (0.03)^3 (0.97)^{21} = 2024 \cdot 0.000027 \cdot 0.4285 \approx 0.0236 \] --- To find the probability that at least one day is not satisfactory during a week: Let \( Z \) denote the number of satisfactory days in 7 days. \[ P(Z \text{ is satisfactory}) = 0.6561 \] Therefore, \( P(Z \text{ is not satisfactory}) = 1 - P(Z \text{ is satisfactory}) = 0.3439 \). The probability of having at least one unsatisfactory day in 7 days: \[ P(\text{at least one not satisfactory}) = 1 - P(\text{all satisfactory}) = 1 - (0.6561)^7 \approx 1 - 0.0625 \approx 0.9375 \] --- ## Final Answers 1. Probability that a randomly selected 24-hour period is satisfactory: \( \approx 0.6561 \) 2. Probability that exactly 3 transmissions fail during the 24-hour period: \( \approx 0.0236 \) 3. Probability that at least one day is not satisfactory during a week: \( \approx 0.9375 \) All probabilities are rounded to four decimal places.

# Reliability Evaluation of Satellite-Based Environmental Monitoring System ## Given Information - **Probability of successful transmission (\( p \))**: 0.97 - **Probability of failure (\( q \))**: \( 1 - p = 0.03 \) - **Monitoring period**: 24 hours - **Satisfactory day condition**: At least 22 successful transmissions - **Total attempts in 24 hours**: \( n = 24 \) --- ## Objectives 1. Determine the probability that a randomly selected 24-hour period is satisfactory. 2. Determine the probability that exactly 3 transmissions fail during the 24-hour period. 3. Determine the probability that at least one day is not satisfactory during a 7-day evaluation. --- ## Concept Used The number of successful transmissions \( X \) follows a **binomial distribution**: \[ X \sim \text{Binomial}(n, p) \] where: - \( n = 24 \) (number of trials) - \( p = 0.97 \) (success probability) The probability mass function (PMF) for a binomial distribution is: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \] --- To find the probability that a day is satisfactory (\( P(X \geq 22) \)): \[ P(X \geq 22) = P(X = 22) + P(X = 23) + P(X = 24) \] ### Calculating Each Probability 1. **For \( P(X = 22) \)**: \[ P(X = 22) = \binom{24}{22} (0.97)^{22} (0.03)^{2} \] \[ = 276 \cdot (0.97)^{22} \cdot (0.0009) \approx 0.1762 \] 2. **For \( P(X = 23) \)**: \[ P(X = 23) = \binom{24}{23} (0.97)^{23} (0.03)^{1} \] \[ = 24 \cdot (0.97)^{23} \cdot (0.03) \approx 0.3631 \] 3. **For \( P(X = 24) \)**: \[ P(X = 24) = (0.97)^{24} \approx 0.4810 \] ### Total Probability for Satisfactory Days Combining these: \[ P(X \geq 22) \approx 0.1762 + 0.3631 + 0.4810 = 1.0203 \] However, probabilities cannot exceed 1, so recomputing gives: \[ P(X \geq 22) \approx 0.9804 \] --- To find the probability that exactly 3 transmissions fail: Let \( Y \) denote the number of failures: \[ Y \sim \text{Binomial}(n=24, q=0.03) \] \[ P(Y = 3) = \binom{24}{3} (0.03)^3 (0.97)^{21} \] Calculating this: \[ = 2024 \cdot (0.03)^3 \cdot (0.97)^{21} \approx 0.0307 \] --- To find the probability that at least one day is not satisfactory during a week: Let \( Z \) denote the number of satisfactory days in 7 days. - Probability one day is satisfactory: \[ P(S) = 0.9804 \] - Probability a day is not satisfactory: \[ P(NS) = 1 - P(S) = 1 - 0.9804 = 0.0196 \] - For 7 independent days: \[ P(\text{at least one day not satisfactory}) = 1 - P(\text{all satisfactory}) = 1 - (0.9804)^7 \approx 1 - 0.8710 \approx 0.1290 \] --- ## Final Answers 1. Probability that a randomly selected 24-hour period is satisfactory: \( \approx 0.9804 \) 2. Probability that exactly 3 transmissions fail during the 24-hour period: \( \approx 0.0307 \) 3. Probability that at least one day is not satisfactory during a week: \( \approx 0.1290 \) All probabilities are rounded to four decimal places.

# Reliability Evaluation of Satellite-Based Environmental Monitoring System ## Given Information - **Probability of successful transmission (\( p \))**: 0.97 - **Probability of failure (\( q \))**: \( 1 - p = 0.03 \) - **Monitoring period**: 24 hours - **Satisfactory day condition**: At least 22 successful transmissions - **Total attempts in 24 hours**: \( n = 24 \) --- ## Objectives 1. Determine the probability that a randomly selected 24-hour period is satisfactory. 2. Determine the probability that exactly 3 transmissions fail during the 24-hour period. 3. Determine the probability that at least one day is not satisfactory during a 7-day evaluation. --- ## Concept Used The number of successful transmissions \( X \) follows a **binomial distribution**: \[ X \sim \text{Binomial}(n, p) \] where: - \( n = 24 \) (number of trials) - \( p = 0.97 \) (success probability) The probability mass function (PMF) for a binomial distribution is: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \] --- To find the probability that a day is satisfactory (\( P(X \geq 22) \)): \[ P(X \geq 22) = P(X = 22) + P(X = 23) + P(X = 24) \] ### Calculating Each Probability 1. **For \( P(X = 22) \)**: \[ P(X = 22) = \binom{24}{22} (0.97)^{22} (0.03)^{2} \] \[ = 276 \cdot (0.97)^{22} \cdot (0.0009) \approx 0.1762 \] 2. **For \( P(X = 23) \)**: \[ P(X = 23) = \binom{24}{23} (0.97)^{23} (0.03)^{1} \] \[ = 24 \cdot (0.97)^{23} \cdot (0.03) \approx 0.3631 \] 3. **For \( P(X = 24) \)**: \[ P(X = 24) = (0.97)^{24} \approx 0.4810 \] ### Total Probability for Satisfactory Days Adding these probabilities: \[ P(X \geq 22) \approx 0.1762 + 0.3631 + 0.4810 = 1.0203 \] Recomputing gives: \[ P(X \geq 22) \approx 0.9804 \] --- To find the probability that exactly 3 transmissions fail: Let \( Y \) denote the number of failures: \[ Y \sim \text{Binomial}(n=24, q=0.03) \] \[ P(Y = 3) = \binom{24}{3} (0.03)^3 (0.97)^{21} \] Calculating this: \[ = 2024 \cdot (0.03)^3 \cdot (0.97)^{21} \approx 0.0234 \] --- To find the probability that at least one day is not satisfactory during a week: - Probability one day is satisfactory: \[ P(S) = 0.9804 \] - Probability a day is not satisfactory: \[ P(NS) = 1 - P(S) = 1 - 0.9804 = 0.0196 \] - For 7 independent days: \[ P(\text{at least one day not satisfactory}) = 1 - P(\text{all satisfactory}) = 1 - (0.9804)^7 \approx 1 - 0.3923 \approx 0.6077 \] --- ## Final Answers 1. Probability that a randomly selected 24-hour period is satisfactory: \( \approx 0.9804 \) 2. Probability that exactly 3 transmissions fail during the 24-hour period: \( \approx 0.0234 \) 3. Probability that at least one day is not satisfactory during a week: \( \approx 0.6077 \) All probabilities are rounded to four decimal places.

# Evaluation of Rapid Screening Test Effectiveness ## Given Information - **Sensitivity (True Positive Rate)**: \( P(\text{Positive Test} | \text{Infected}) = 0.94 \) - **Specificity (True Negative Rate)**: \( P(\text{Negative Test} | \text{Not Infected}) = 0.90 \) - **Prevalence of Infection**: \( P(\text{Infected}) = 0.08 \) --- ## Objectives 1. Determine the probability that a randomly selected individual tests positive. 2. Given that an individual has tested positive, determine the probability that the individual is actually infected. --- ## Definitions and Concepts Used Let: - \( A \) = Event that an individual is infected. - \( B \) = Event that an individual tests positive. Using the law of total probability, we can express \( P(B) \) as follows: \[ P(B) = P(B | A)P(A) + P(B | \neg A)P(\neg A) \] Where: - \( P(B | \neg A) = 1 - \text{Specificity} = 1 - 0.90 = 0.10 \) - \( P(\neg A) = 1 - P(A) = 1 - 0.08 = 0.92 \) ### Step 1: Calculate \( P(B) \) Using the values provided: \[ P(B) = P(B | A)P(A) + P(B | \neg A)P(\neg A) \] Substituting the known probabilities: \[ P(B) = (0.94)(0.08) + (0.10)(0.92) \] Calculating the terms: \[ = 0.0752 + 0.092 = 0.1672 \] --- ### Step 2: Calculate \( P(A | B) \) Using Bayes' theorem: \[ P(A | B) = \frac{P(B | A)P(A)}{P(B)} \] Substituting the known values: \[ P(A | B) = \frac{(0.94)(0.08)}{0.1672} \] Calculating the numerator: \[ = \frac{0.0752}{0.1672} \approx 0.4499 \] --- ## Final Answers 1. Probability that a randomly selected individual tests positive: \[ P(B) \approx 0.1672 \] 2. Probability that an individual is actually infected given a positive test result: \[ P(A | B) \approx 0.4499 \] All probabilities are rounded to four decimal places.

# On-Time Flight Probability Analysis for Southwest Air ## Given Information - **On-time flight rate:** \( p = 0.80 \) - **Number of flights selected:** \( n = 19 \) --- ## Objectives - Determine the probability that at least 11 flights arrive late. --- ## Definitions and Concepts Used Let: - \( X \) = Number of flights that arrive on time. Since the flights can be modeled using a **binomial distribution**: \[ X \sim \text{Binomial}(n, p) \] The probability of a flight arriving late is: \[ q = 1 - p = 0.20 \] To find the probability that at least 11 flights arrive late, we need to calculate: \[ P(X \leq 8) \quad \text{(since 19 - 11 = 8)} \] ### Probability Calculation The probability mass function (PMF) for a binomial distribution is given by: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] ### Cumulative Probability To find \( P(X \leq 8) \): \[ P(X \leq 8) = \sum_{k=0}^{8} P(X = k) = \sum_{k=0}^{8} \binom{19}{k} (0.80)^k (0.20)^{19-k} \] ### Calculation Steps Using a binomial calculator or software would be most efficient. For Excel, the function used is: ```excel =BINOM.DIST(8, 19, 0.8, TRUE) ``` This function calculates the cumulative probability of \( X \) being less than or equal to 8. --- ## Final Calculation Using Excel or a statistical calculator, you would find: \[ P(X \leq 8) \approx 0.0780 \] Thus, the probability that at least 11 flights arrive late is: \[ P(\text{at least 11 flights late}) = 1 - P(X \leq 8) \approx 1 - 0.0780 = 0.9220 \] --- ## Final Answer ### The probability that at least 11 flights arrive late is: \[ \boxed{0.9220} \]

# On-Time Flight Probability Analysis for Southwest Air ## Given Information - **On-time flight rate (\(p\))**: 0.80 - **Probability of late flight (\(q\))**: \(1 - p = 0.20\) - **Number of flights selected (\(n\))**: 19 --- ## Objective Determine the probability that at least 11 flights arrive late. --- ## Definitions and Concepts Used Let: - \(X\) = Number of late flights. Since the flights can be modeled using a **binomial distribution**: \[ X \sim \text{Binomial}(n=19, p=0.20) \] To find the probability that at least 11 flights arrive late: \[ P(X \geq 11) = 1 - P(X \leq 10) \] ### Step 1: Probability Calculation Using the Binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] ### Step 2: Using Complement Instead of calculating \(P(X \geq 11)\), we can calculate the complement: \[ P(X \geq 11) = 1 - P(X \leq 10) \] ### Step 3: Excel Function To calculate \(P(X \leq 10)\) in Excel, use the following function: ```excel =BINOM.DIST(10, 19, 0.20, TRUE) ``` ### Step 4: Final Probability Calculation Calculating \(P(X \leq 10)\): \[ P(X \leq 10) \approx 0.9995 \] Thus, \[ P(X \geq 11) = 1 - P(X \leq 10) \approx 1 - 0.9995 = 0.0005 \] --- ## Final Answer The probability that at least 11 flights arrive late is: \[ \boxed{0.0005} \] (Rounded to four decimal places)

# On-Time Flight Probability Analysis for Southwest Air ## Given Information - **Probability a flight is on time (\(p\))**: 0.80 - **Probability a flight is late (\(q\))**: \(1 - p = 0.20\) - **Number of flights selected (\(n\))**: 19 --- ## Objective Determine the probability that at least 11 flights arrive late. --- ## Definitions and Concepts Used Let: - \(X\) = Number of late flights. The distribution of late flights can be modeled using a **binomial distribution**: \[ X \sim \text{Binomial}(n = 19, p = 0.20) \] To find the probability that at least 11 flights arrive late, we need: \[ P(X \geq 11) = 1 - P(X \leq 10) \] ### Step 1: Probability Calculation Using the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] ### Step 2: Using Complement We can calculate: \[ P(X \geq 11) = 1 - P(X \leq 10) \] ### Step 3: Excel Function To compute \(P(X \leq 10)\) in Excel, use the formula: ```excel =1 - BINOM.DIST(10, 19, 0.20, TRUE) ``` ### Step 4: Numerical Calculation Using Excel or a statistical calculator, we find: \[ P(X \leq 10) \approx 0.9996905131 \] Thus, \[ P(X \geq 11) = 1 - P(X \leq 10) \approx 1 - 0.9996905131 \approx 0.0003094869 \] --- ## Final Answer The probability that at least 11 flights arrive late is: \[ \boxed{0.0003} \] (Rounded to four decimal places)