1. BCOR 1020 Business Statistics Lecture 13 – February 28, 2008

2. Overview Chapter 7 – Continuous Distributions –Normal Distribution (continued)

3. Chapter 7 – Normal Distribution Recall the Standard Normal: Since for every value of  and , there is a different normal distribution, we transform a normal random variable to a standard normal distribution with  = 0 and  = 1 using the formula: z = x –   Denoted N(0,1) Appendix C-2 allows you to find all of the area under the curve left of z. (Hand-out) Shift the point of symmetry to zero by subtracting  from x. Divide by  to scale the distribution to a normal with  = 1.

4. Chapter 7 – Normal Distribution Example: Using the Std. Normal transformation Daily sales at a bicycle shop are normally distributed with mean $15,000 and standard deviation $4000. Find the probability that sales will exceed $20,000 on a randomly- selected day. Find the probability that sales will be less than $12,000 on a randomly selected day. Find the probability that sales will be between $12,000 and $20,000 on a randomly selected day. P(X > 20000) = 1 – P(X < 20000) P(X < 12000) = P(12000 < X < 20000) = P (X < 20000) – P(X < 12000)

5. Clickers If the starting salary for students majoring in Business is normally distributed with a mean of $45,000 and a standard deviation of $5,000, find the probability that the starting salary of a randomly selected student will be less than $50,000. A = 0.1056 B = 0.3085 C = 0.6915 D = 0.8413

6. Clickers If the starting salary for students majoring in Business is normally distributed with a mean of $45,000 and a standard deviation of $5,000, find the probability that the starting salary of a randomly selected student will be at least $50,000. A = 0.1056 B = 0.1587 C = 0.8413 D = 0.8944

7. Clickers If the starting salary for students majoring in Business is normally distributed with a mean of $45,000 and a standard deviation of $5,000, find the probability that the starting salary of a randomly selected student will be between $45,000 and $50,000. A = 0.3413 B = 0.3944 C = 0.8413 D = 0.8944

8. Chapter 7 – Normal Distribution Basis for the Empirical Rule: Approximately 68% of the area under the curve is between + 1  Approximately 95% of the area under the curve is between + 2  Approximately 99.7% of the area under the curve is between + 3 

9. Chapter 7 – Normal Distribution Finding z for a Given Area: Appendices C-1 and C-2 be used to find the z-value corresponding to a given probability. For example, what z-value defines the top 1% of a normal distribution? This implies that 99% of the area lies to the left of z. Or that 1% of the area lies to the left of –z. Or that 49% of the area lies between 0 and z.

10. Chapter 7 – Normal Distribution Finding z for a Given Area: Look for an area of.4900 in Appendix C-1 (or for an area of 0.9900 in the Appendix C-2 – the handout): Without interpolation, the closest we can get is z = 2.33

11. Chapter 7 – Normal Distribution Finding z for a Given Area: Some important Normal areas:

12. Chapter 7 – Normal Distribution Finding Areas by Standardizing: Suppose John took an economics exam on which the class mean was 75 with a standard deviation of 7. What score would place John in the upper 10 th percentile? Find the value of x such that P(X > x) =.10 or P(X < x) = 0.90. From the previous slide, we know P(Z > 1.282) =.10. Since A score of 84 or better would place John in the top 10% of his class.

13. Clickers Suppose the starting salary for students majoring in Business is normally distributed with a mean of $45,000 and a standard deviation of $5,000. If Jane Wants a starting salary in the top 25%, approximately what salary should she negotiate for? A = $56,630 B = $54,800 C = $53,225 D = $51,410 E = $48,375

14. Chapter 7 – Normal Distribution Normal Probabilities and z-scores in MegaStat (time allowing)