FRANKIE AMOS-F O O D S E R V I C E M A N A G E M E N T

FRIDA BECKMAN- LEHISTE -S P O R T S / E V E N T / E N T E R TA I N M E N T

M A N A G E M E N T

JOHN TEJADA -M A N A G E M E N T / B U S I N E S S

A D M I N I S T R AT I O N

ROCCO LAVAGLIO -M A N A G E M E N T

Stats Bar-n-Grill

Tonight’s MenuOrganizing Data- Rocco

Basic Probability-Frida

Probability & Binomial Distributions-John

Regression & Correlation-Frankie

Stats Bar n Grill

I. Lesson 2

Class Frequency (f)

Relative Frequenc

y

Cumulative

Frequency

Class Mark

Boundaries

menu prices f/n midpoint

$2.75-$4.95 25 0.125 25 $3.85 $2.745-$4.955$4.96-$7.15 50 0.25 75 $6.05 $4.955-$7.155$7.16-$9.35 38 0.19 113 $8.25 $7.155-$9.355$9.36.$11.55 45 0.225 158 $10.45 $9.355-$11.555$11.56-$13.75

10 0.05 168 $12.65 $11.555-$13.755

$13.76-$15.95

15 0.075 183 $14.85 $13.755-$15.955

$15.96-$18.15

12 0.06 195 $17.05 $15.955-$18.155

$18.16-$18.95

5 0.025 200 $18.55 $18.155-18.955

n=200 1

II. Lessons 8,9,10,11

Valet Parking experimentExperimental (Imperial) ProbabilityOutcome 0≤P(Event)≥1Study for November: n=312 cars

event= 22 accidents

P(accident)= 22÷312= 0.07=7%P(accident)’=290÷312= 0.93=93%

II. Lessons 8,9,10,11

How can our restaurant use these statistics? Insurance rates Keep track of how the probability is changing Advertise the valet parking

Different use of statistics and why you always should think twice:

93% non-accident rate sounds great7% chance that your car is involved in an accident does

sound worseHave to question how big was the sample size, over what

period of time etc. To determine the credibility.

II. Lessons 8,9,10,11

Day Scratch Dent Marginal TotalMonday 0 0 0Tuesday 1 0 1

Wednesday 2 1 3Thursday 3 2 5

Friday 4 3 7Saturday 3 2 5Sunday 1 0 1

Marginal Total 14 8 22

Looking at the 22 cars involved in accidents over the month and how they were spread through the week days in a CONTINGENCY TABLE:

II. Lessons 8,9,10,11

Addition Rule:P(AorB)=P(A)+P(B)-P(A∩B) P(Scratch)= (14÷22)= 0.64=64% P (Dent)= (8÷22)= 0.36= 36% P(Scratch or Dent)= (14÷22)+(8÷22)=22÷22= 1=100%Mutually exclusive P(Friday or Dent)= (7÷22)+(8÷22)-(3÷22)= 12÷22=0.55=55% P(Wednesday or Scratch)= (3÷22)+(14÷22)-(2÷22)=

15÷22=0.68=68%Not mutually exclusive

II. Lessons 8,9,10,11

Conditional probability-given informationP(Dent|Saturday)= (2|5)= 2÷5=0.4=40%P(Sunday|Scratch)= (1|1)= 1÷1=1=100%Multiplication Rule-more than one selection• Choose 2 daysP(Fri. and Sat.)=(7÷22)×(5÷22)= 12÷22=0.55=55%P(Mon and Tue)=(0÷22)×(1÷22)=1÷22=0.045=4.5%

Day Scratch Dent Marginal TotalMonday 0 0 0Tuesday 1 0 1

Wednesday 2 1 3Thursday 3 2 5

Friday 4 3 7Saturday 3 2 5Sunday 1 0 1

Marginal Total 14 8 22

II. Lessons 8,9,10,11

How can the restaurant use the weekly table and the probability generated from that?

Insurance purposesLegal purposesManager evaluationsSpaceStaffing

III. Lesson 19

Applications of the Normal Distributionμ 23.58 σ 3.68 What is the probability that

a patron to Stat’s Bar & Grill will spend an amount between $25.00 and $33.00 on a meal?

III. Lesson 19

Cont. Applications of the Normal DistributionLet x be the random variable which represents

how much a patron spends on his meal:P(25<x<33)

Remember the z score formula:

Z=X-μ σ

Cont. Applications of the Normal Distribution

Z score for $25.00

(25-23.58)/3.68=0.39

Z Table= .1517

Z score for $33.00

(33-23.58)/3.68=2.56

Z Table Value= .4948

III. Lesson 19

III. Lesson 19

Cont. Applications of the Normal Distribution

.4948- .1517=.3431

There is a 34.31% probability that a patron will spend an amount between $25.00 and $33.00 on a meal at Stats Bar & Grill.

IV. Lessons 30, 31

Correlation and Regression

Determine the amount of staff needed daily.Discover minimum staffing needs for special

functions.

Does the amount of customers influence the staffing needs?

Is it a Positive or Negative correlation?

Average Customer CountSunda

yMonday Tuesda

yWednesday Thursda

yFriday Saturda

yCustome

rs170 140 210 190 240 290 360

Staff 5 4 6 5 7 8 10

3 4 5 6 7 8 9 10 110

50100150200250300350400

x y ∑xy ∑x² ∑y²5 170 850 25 289004 140 560 16 196006 210 1260 36 441005 190 950 25 361007 240 1680 49 576008 290 2320 64 8410010 360 3600 100 129600= 45 = 1600 = 11220 = 315 = 400000

High Positive Correlation

3 4 5 6 7 8 9 10 11050

100150200250300350400

f(x) = 36.3333333333333 x − 5

IV. Lessons 31, 31

Thank You for listening!

Any Questions?