Question

Consider the probability distribution of the weekly demand for copier paper (in hundreds of reams) used in a corporation's duplicating center, as shown in the file P04_27.xlsx.

a. Use simulation to generate 500 values of this random variable. Calculate the mean and standard deviation of the given distribution. (Round your answers to two decimal places.)

Mean
Standard Deviation

b. Find the mean and standard deviation of the simulated values. (Round your answers to two decimal places.)

Mean
Standard Deviation

c. Use your simulated values to estimate the probability that weekly copier paper demand is within one standard deviation of the mean. (Round your answers up to one decimal place.)

Mean - Standard Deviation
Mean + Standard Deviation
P(demand within 1 Std. Dev. of the Mean)	%

Why is this only an estimate, not an exact value?

Here is the info in the excel sheet

Weekly copier paper demand
Demand	Probability
10	0.05
11	0.10
12	0.14
13	0.16
14	0.21
15	0.13
16	0.11
17	0.04
18	0.03
19	0.02
20	0.01

Answer 1

Demand Probability Cumulative Probability 10 0.05 12 13 14 15 16 17 18 19 20 0.14 0.16 0.21 0.13 0.11 0.04 0.03 0.02 0.01 0.05 0.15 0.29 0.45 0.66 0.79 0.9 0.94 0.97 0.99

I am unable to show all the 500 simulations here, so I am showing you the first 100 only.

Random #	Demand
0.2853049	12
0.1314233	11
0.7103724	15
0.7141029	15
0.0424316	10
0.0432327	10
0.6624186	15
0.6974529	15
0.2928261	13
0.5127866	14
0.7191688	15
0.8743126	16
0.5971292	14
0.8753781	16
0.7053927	15
0.2994464	13
0.1472498	11
0.3052605	13
0.7218783	15
0.9778462	19
0.9357838	17
0.6586554	14
0.532113	14
0.2627841	12
0.3621545	13
0.8058528	16
0.6430508	14
0.4532812	14
0.5949728	14
0.9100599	17
0.6525201	14
0.6520141	14
0.7329046	15
0.6616399	15
0.0692224	11
0.3073655	13
0.9700517	19
0.0719604	11
0.1973078	12
0.1231753	11
0.26169	12
0.8621058	16
0.449521	13
0.133875	11
0.3841196	13
0.6939706	15
0.8619227	16
0.6258181	14
0.7355935	15
0.6808148	15
0.3317924	13
0.5110906	14
0.9698784	18
0.569545	14
0.37036	13
0.9515356	18
0.1448155	11
0.3119094	13
0.3153497	13
0.320733	13
0.1451369	11
0.7896199	15
0.5518004	14
0.3953913	13
0.4387107	13
0.3048158	13
0.4857281	14
0.9524587	18
0.8009903	16
0.620837	14
0.3953379	13
0.4910639	14
0.4067192	13
0.5164161	14
0.0533258	11
0.8263863	16
0.0622823	11
0.7067224	15
0.8678687	16
0.6119928	14
0.8139332	16
0.3917536	13
0.9032413	17
0.1905865	12
0.4529227	14
0.4640331	14
0.4241396	13
0.8264788	16
0.718925	15
0.0985041	11
0.0372518	10
0.8009729	16
0.1876476	12
0.3021737	13
0.5910082	14
0.6032203	14
0.2272108	12
0.8265451	16
0.548562	14
0.5501075	14

Mean, μ = 13.85 (b)        

Standard deviation, σ = 2.16      

(c)        

μ - σ = 11.69       

μ + σ = 16.02       

Number of simulations where the demand lies in the range [11.69, 16.02] = 372

P(11.69 < Demand < 16.02) = 0.7     

(d) This is only an estimate and not an exact value because this is based on just 500 simulations.

[Please give me a Thumbs Up if you are satisfied with my answer. If you are not, please comment on it, so I can edit the answer. Thanks.]

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