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antoniya [11.8K]
3 years ago
14

A random sample of 28 items is drawn from a population whose standard deviation is unknown. The sample mean is x=900 and the sam

ple deviation is s = 10. Use appendix D to find the values of student's t. (a) Construct an interval estimate of u with 98% confidence. (Round your answers to 3 decimal places.)
Mathematics
1 answer:
tatiyna3 years ago
7 0

Answer:

<em>795.596≤u≤804.403</em>

Step-by-step explanation:

Confidence interval is expressed using the formula;

CI = xbar ± (z*s/√n)

xbar is the sample mean

z is the z score at 98% confidence

s is the standard deviation

n is the sample size

Given

xbar = 800

z = 2.33

s = 10

n = 28

Substitute into the formula;

CI =800 ± (2.33*10/√28)

CI = 800 ± (2.33*10/5.2915)

Ci = 800± (2.33* 1.8898)

CI = 800 ± 4.4033

CI = (800-4.4033, 800 + 4.4033)

CI = (795.596, 804.403)

<em>Hence an interval estimate of u with 98% confidence is expressed as </em>

<em>795.596≤u≤804.403</em>

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Given Information:

Average customer arrival = λ =2.3 customers/hour

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Required  Information:

Use one consultant with an average service time of 8 minutes/customer or use two consultants, each with average service time of 10 minutes/customer ?

Answer:

it is recommended to use one consultant with an average service time of 8 minutes/customer.

Step-by-step explanation:

Convert the average consultant time to hours/customer

Average consultant time1 = μ  = 60/8 = 7.5 hours/customer

Average consultant time2 = μ  = 60/10 = 6 hours/customer

Single consultant queuing model:

Calculations for one consultant

Average no. of customers waiting for service = Lq = λ²/μ(μ-λ)

Average no. of customers waiting for service = Lq = (2.3)²/7.5(7.5 - 2.5)

Average no. of customers waiting for service = Lq = 0.1356

Average no. of customers in the system = L = Lq + λ/μ

Average no. of customers in the system = L = 0.1356 + 2.3/7.5

Average no. of customers in the system = L = 0.442

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Total cost = $25*0.442 + $16

Total cost = $27.05

Multi consultant queuing model:

Calculations for two consultants

Average no. of customers waiting for service = Lq = ((λ/μ)^k(λμ)/(k -1)(kμ - λ))*P₀

Where k = 2 is the number of consultants and P₀ is the probability that all of the k consultants are idle. The value of P₀  can be found in the tables with λ/μ = 2.3/6 = 0.38 and k = 2,  P₀ ≅ 0.685

Average no. of customers waiting for service = Lq = (2.3/6)²(2.3*6)/(2-1)(2*6 - 2.3)*0.685

Average no. of customers waiting for service = Lq = 0.1431

Average no. of customers in the system = L = Lq + λ/μ

Average no. of customers in the system = L = 0.1431 + 2.3/6

Average no. of customers in the system = L = 0.5264

Total cost = (customer waiting time cost)*L + consultant service cost

Total cost = $25*0.5264 + ($16)*2 (since there are 2 consultants now)

Total cost = $45.16

Conclusion:

Therefore, it is recommended to use one consultant with an average service time of 8 minutes per customer since the total cost is less than the other option.

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