1. Application of Standard Normal Distribution Q1: A certain type of storage battery lasts, on average 3.0 years with standard deviation of 0.5 year, assuming that battery life is normally distributed; find the probability that a given battery will last less than 2.3 years.

2. Application of Standard Normal Distribution

3. Application of Standard Normal Distribution Solution: We want to find P( X < 2.3) to find this area or probability by computing z as P(X < 2.3) = P( z < -1.4 ) From the normal distribution table we can conclude that the probability is 0.0808

4. Application of Standard Normal Distribution Q2: An electrical manufactures light bulbs that have a life before burn out that is normally distributed with mean equal to 800 hours and a standard deviation of 40 hours. Find the probability that a bulb burns between 778 and 834 hours.

5. Application of Standard Normal Distribution Solution: We should calculate z 1 and z 2 as:

6. Application of Standard Normal Distribution Then P (778 < X< 834) = P (z < 0.85) – P (z < -0.55) From table P (z < 0.85) – P (z < -0.55) = 0.8023 – 0.2912 = 0.5111

7. Application of Standard Normal Distribution Q3 : In an industrial process, the diameter of a ball bearing is an important measurement. The buyer sets specifications for diameter to be. The implication is that no part falling outside these specifications will be accepted. It is known that in the process the diameter of a ball bearing has a normal distribution with mean = 3.0 and standard deviation = 0.005. On average, how many manufactured ball bearing will be scrapped?

8. Application of Standard Normal Distribution Solution:

9. Application of Standard Normal Distribution P (2.99 < X < 3.01) = P (-2.0 < Z < +2) P (-2.0<Z)= 0.0228 P (Z < +2) = 1 - P (+2<Z) = 1- 0.9772 = 0.0228 Then the average that is So that 4.56% are scrapped

10. Application of Standard Normal Distribution Q4: Gauges are used to reject all components for which a certain dimension is not within the specification. It is known that this measurement is normally distribution with mean 1.50 and standard deviation 0.2. Determine the value d such that the specifications cover 95% of the measurements.

11. Application of Standard Normal Distribution The left area = 100 – 95 = 5% for both sides

12. Application of Standard Normal Distribution The area is equal to 1 – 0.025 = 0.975 From table this area equal to z1 or z2 = 1.96 Or

13. Application of Standard Normal Distribution Q5: A certain machine makes electrical resisters having mean resistance of 40 ohms and standard deviation of 2.0 ohms. Assuming that the resistance follows a normal distribution and can be measured to any degree of accuracy, what percentage of resistors will have resistance exceed 43 ohms?

14. Application of Standard Normal Distribution Solution: P (X > 43) = P(Z > 1.5) = 1- P(Z < 1.5) =1 – 0.9332=0.0668

15. Application of Standard Normal Distribution Solution: Then the percentage of resistors will have resistance exceed 43 ohms = 6.68%

16. Application of Standard Normal Distribution Q6: The average grade of an exam is 74% and the standard deviation is 7.0. If 12% of the class is given A’s and the grades are curved to follow normal distribution, what is the lowest possible A and the highest possible B?

17. Application of Standard Normal Distribution Solution: The area left 12% is 88% = 0.88 From table we found the z value is 1.18

18. Application of Standard Normal Distribution Then Lowest A is 82 and highest B is 83