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THIS MUST BE IN EXCEL... SHOW FORMULAS USED Problem 1: On the average 10 custome

ID: 3367358 • Letter: T

Question

THIS MUST BE IN EXCEL... SHOW FORMULAS USED

Problem 1: On the average 10 customers arrive each half-an-hour at the checkout counter at a local retail store.

Part A: Prepare the probability distribution table in Excel. What is the probability that more than 8 customers will arrive at the checkout counter in 24 min.?

Part B: Prepare the probability distribution table in Excel. What is the probability that less than 6 customers arrive at the checkout counter in 27 min.?

Part C: Prepare the probability distribution table in Excel. What is the probability that no more than 7 customers arrive at the checkout counter in 30 min.?

Part D: Prepare the probability distribution table in Excel. What is the probability that no less than 3 customers arrive at the checkout counter in 18 min.?

Part E: Prepare the probability distribution table in Excel. What is the probability that at least 4 but less than 10 customers arrive at the checkout counter in 22.5 min.?

Explanation / Answer

Solution

Let X = Number of customers who arrive each half-an-hour at the checkout counter at a local retail store. Then, X ~ Poisson (?), where ? = 10 [given] ……………………(A)

Back-up Theory

If a random variable X ~ Poisson(?), i.e., X has Poisson Distribution with mean ? then

probability mass function (pmf) of X is given by P(X = x) = e – ?.?x/(x!) …………..(1)

where x = 0, 1, 2, ……. , ?

Values of p(x) for various values of ? and x can be obtained by using Excel Function, POISSON(x,Mean,Cumulative)

If X = number of times an event occurs during period t, Y = number of times the same event occurs during period kt, and X ~ Poisson(?), then Y ~ Poisson (k?) ……….. (2)

Part (a)

Let X = Number of customers who arrive each 24 minutes at the checkout counter at a local retail store. Then, vide (2), X ~ Poisson (8), [ ? = 10 for 30 minutes => 8 for 24 minutes]. Using Excel Function, POISSON(x,Mean,Cumulative), probability distribution of X is given below:

? = 8; p(x) = P(X = x); P(x) = ?p(x)

x

0

1

2

3

4

5

6

7

8

p(x)

0.0003

0.0027

0.0107

0.0286

0.0573

0.0916

0.1221

0.1396

0.1396

P(x)

0.0003

0.0030

0.0138

0.0424

0.0996

0.1912

0.3134

0.4530

0.5925

So, probability more than 8 customers arrive in 24 minutes at the checkout counter

= 1 - ?p(x)(x = 0 to 8)

= 1 – 0.5925

= 0.4075 ANSWER

Part (b)

Let X = Number of customers who arrive each 27 minutes at the checkout counter at a local retail store. Then, vide (2), X ~ Poisson (9), [ ? = 10 for 30 minutes => 9 for 27 minutes]. Using Excel Function, POISSON(x,Mean,Cumulative), probability distribution of X is given below:

? = 9; p(x) = P(X = x); P(x) = ?p(x)

x

0

1

2

3

4

5

p(x)

0.0001234

0.00111069

0.004998

0.014994

0.033737

0.060727

P(x)

0.0001234

0.0012341

0.006232

0.021226

0.054964

0.115691

So, probability less than 6 customers arrive in 27 minutes at the checkout counter

= ?p(x)(x = 0 to 5)

= 0.1157 ANSWER

Part (c)

Let X = Number of customers who arrive each 30 minutes at the checkout counter at a local retail store. Then, vide (2), X ~ Poisson (10), [ ? = 10 for 30 minutes is given]. Using Excel Function, POISSON(x,Mean,Cumulative), probability distribution of X is given below:

? = 10; p(x) = P(X = x); P(x) = ?p(x)

x

0

1

2

3

4

5

6

7

p(x)

4.54E-05

0.0005

0.0023

0.0076

0.0189

0.0378

0.0631

0.0901

P(x)

4.54E-05

0.0005

0.0028

0.0103

0.0293

0.0671

0.1301

0.2202

So, probability no more than 7 customers arrive in 24 minutes at the checkout counter

= ?p(x)(x = 0 to 7)

= 0.2202 ANSWER

Part (d)

Let X = Number of customers who arrive each 18 minutes at the checkout counter at a local retail store. Then, vide (2), X ~ Poisson (6), [ ? = 10 for 30 minutes => 6 per 18 minutes]. Using Excel Function, POISSON(x,Mean,Cumulative), probability distribution of X is given below:

? = 6; p(x) = P(X = x); P(x) = ?p(x)

x

0

1

2

3

p(x)

0.0024788

0.01487251

0.044618

0.089235

P(x)

0.0024788

0.01735127

0.061969

0.151204

So, probability no more than 3 customers arrive in 18 minutes at the checkout counter

= ?p(x)(x = 0 to 3)

= 0.1512 ANSWER

Part (e)

Let X = Number of customers who arrive each 22.5 minutes at the checkout counter at a local retail store. Then, vide (2), X ~ Poisson (7.5), [ ? = 10 for 30 minutes => 7.5 per 22.5 minutes]. Using Excel Function, POISSON(x,Mean,Cumulative), probability distribution of X is given below:

? = 7.5; p(x) = P(X = x); P(x) = ?p(x)

x

0

1

2

3

4

5

6

7

8

9

p(x)

0.0006

0.0041

0.0156

0.0389

0.0729

0.1094

0.1367

0.1465

0.1373

0.1144

P(x)

0.0006

0.0047

0.0203

0.0591

0.1321

0.2414

0.3782

0.5246

0.6620

0.7764

So, probability at least 4, but less than 10 customers arrive in 22.5 minutes at the checkout counter

= ?p(x)(x = 4 to 9)

= ?p(x)(x = 0 to 9) - ?p(x)(x = 0 to 3)

= 0.7764 – 0.0591

= 0.7173 ANSWER

DONE

x

0

1

2

3

4

5

6

7

8

p(x)

0.0003

0.0027

0.0107

0.0286

0.0573

0.0916

0.1221

0.1396

0.1396

P(x)

0.0003

0.0030

0.0138

0.0424

0.0996

0.1912

0.3134

0.4530

0.5925