Standard Normal Distribution Part II The standard normal distribution is continuous.

Standard Normal Distribution Part II The standard normal distribution is continuous. Z-score formula to solve for probability from x-values using the standard normal distribution: (x-4) z= We have used the formula above to go from x-values to 2-scores to probabilities. However, if we know a distribution is normal and we know the mean and standard deviation, we can go in the opposite direction as wel… probabilities to z-scores to x-values. Consider the Meadas shocks example by plugging in the mean and standard deviation into the above formula: (1-60,00) z = 10,000 So, if we have an x-value we can solve for a Z-score. Or, if we have a Z-score we can do a little algebra and solve for an x-value To retrieve a 2-score from a cumulative probability: =norm.s.inv(probability) Note: You may need to convert the 2-score to an x-value to solve the problem you are working on. Meadas company makes shocks. The distance traveled before the shocks wear out is normally distributed with population mean = 60,000 miles and population standard deviation = 10,000 miles. 11. What is the 2-score for cumulative probability = 0.207 12. At what number of miles is there a 0.20 probability the shock will have failed? 13. What is the 2-score for cumulative probability = 0.8? 14. At what number of miles is there a 80% chance that the shock will have previously failed? Return to the golf club example with the original and new driver. Assume that the original driver has population mean = 280 yards and population standard deviation – 15 yards. 15. What is P(x < 280)=”” for=”” the=”” original=”” driver?=”” 16.=”” at=”” what=”” number=”” of=”” yards=”” is=”” there=”” a=”” 5%=”” chance=”” that=”” a=”” drive=”” would=”” be=”” a=”” shorter=”” distance=”” with=”” the=”” original=”” driver?=”” 17.=”” at=”” what=”” number=”” of=”” yards=”” is=”” there=”” a=”” 95%=”” chance=”” that=”” a=”” drive=”” would=”” be=”” a=”” shorter=”” distance=”” with=”” the=”” original=”” driver?=”” assume=”” the=”” new=”” driver=”” has=”” population=”” mean=”290″ yards=”” and=”” population=”” standard=”” deviation=”17″ yards.=”” 18.=”” what=”” is=””><>< 300)=”” for=”” the=”” new=”” driver?=”” 19.=”” at=”” what=”” number=”” of=”” yards=”” is=”” there=”” a=”” 2.5%=”” chance=”” that=”” a=”” drive=”” would=”” be=”” a=”” shorter=”” distance=”” with=”” the=”” new=”” driver?=”” 20.=”” at=”” what=”” number=”” of=”” yards=”” is=”” there=”” a=”” 97.5%=”” chance=”” that=”” a=”” drive=”” would=”” be=”” a=”” shorter=”” distance=”” with=”” the=”” new=”” driver?=””><>

 


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