BCOR 1020Business Statistics

Lecture 13 – February 28, 2008

Overview

• Chapter 7 – Continuous Distributions– Normal Distribution (continued)

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.

Chapter 7 – Normal DistributionExample: 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.

1056.8944.1)25.1(1)(1 40001500020000 ZPZP

P(X > 20000) = 1 – P(X < 20000)

P(X < 12000) = 2266.)75.0()( 40001500012000 ZPZP

P(12000 < X < 20000) = P (X < 20000) – P(X < 12000)

6678.2266.8944.)75.0()25.1( ZPZP

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

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

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

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

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.

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

Chapter 7 – Normal Distribution

Finding z for a Given Area:• Some important Normal areas:

Chapter 7 – Normal Distribution

, xz 282.17

75x

974.83 x

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 10th 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 974.875x

• A score of 84 or better would place John in the top 10% of his class.

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

Chapter 7 – Normal Distribution

• Normal Probabilities and z-scores in MegaStat (time allowing)