Question

1- KALI, Inc., manufactures home appliances that are marketed under a variety of trade names. However, KALI does not manufacture every component used in its products. Several components are purchased directly from suppliers. For example, one of the components that KALI purchases for use in home air conditioners is an overload protector, a device that turns off the compressor if it overheats. The compressor can be seriously damaged if the overload protector does not function properly, and therefore KALI is concerned about the quality of the overload protectors. One way to ensure quality would be to test every component received through an approach known as 100% inspection. However, to determine proper functioning of an overload protector, the device must be subjected to time-consuming and expensive tests, and KALI cannot justify testing every overload protector it receives.

Instead, KALI uses an acceptance sampling plan to monitor the quality of the overload protectors. The acceptance sampling plan requires that KALI's quality control inspectors select and test a sample of overload protectors from each shipment. If very few defective units are found in the sample, the lot is probably of good quality and should be accepted. However, if a large number of defective units are found in the sample, the lot is probably of poor quality and should be rejected.

The quality control manager requested a producer's risk of 0.10 when

p₀

was 0.05 and a consumer's risk of 0.20 when

p₁

was 0.20. Consider the acceptance sampling plan based on a sample size of 15 and an acceptance number of 1. Answer the following questions.

(a)

What is the producer's risk for the

n = 15, c = 1

sampling plan? (Round your answer to four decimal places.)

(b)

What is the consumer's risk for the

n = 15, c = 1

sampling plan? (Round your answer to four decimal places.)

(c)

Does the

n = 15, c = 1

sampling plan satisfy the risks requested by the quality control manager? Discuss. (Select all that apply.)

The producer's risk is acceptable since it is less than 0.10.The producer's risk is not acceptable since it is greater than 0.10.The consumer's risk is acceptable since it is less than 0.20.The consumer's risk is not acceptable since it is greater than 0.20.

2-

Temperature is used to measure the output of a production process. When the process is in control, the mean of the process is

μ = 123.5

and the standard deviation is

σ = 0.3.

(a)

Compute the upper and lower control limits if samples of size 6 are to be used. (Round your answers to two decimal places.)

UCL= LCL=

(b)

Consider a sample providing the following data.

123.8

123.2

124.1

123.7

123.4

124.2

Compute the mean for this sample. (Round your answer to two decimal places.)

x

=  

c)

Consider a sample providing the following data.

124.3

123.7

123.6

124.2

124.5

129.0

Compute the mean for this sample. (Round your answer to two decimal places.)

x

=

Answer 1

Pls, comment if anu issue.

a) Hese, n= 15,2=1, PO = 0.05 fcx)= n! a! (n-x)! pl (1-P) - hos Y LO, f(0) = 15! (003) (-0.05)95-0 0!(15-0) 0-46329 こ Forx=1 F C1= 151 Coos)'C-os)'54 !! (15-1)! = 0:36575 The probability that pois defective will he accepted 1- CHF 10.463 297+0-36575) -10.17905 b) Here, n=15,C=1 IP = 0.20 f(x)= n! px Cirpinux x1(m.xd! For y=0, fcola 15! (0-2) C1-0-2)5-4 0!150)! -0.03518 Foo x=1 FW= 15! (0.23' (1-0.211511 " (15-13 = 0.13195 The probability that pris defective will ho accepted ct(0) PFU) = 0.035184 0.13199

10.16712 e) The produres's sisk is not acceptable since it is greater than o. 1 The consumed disk is acceptable since it is less tan 0.2 2) u = 123-5,0=0.3 a) OCLEM + 30 va o 123.5+ 3x0.3 vo = 123.8679 2 LCL - N - 30 =123.5-3x0.3 123-13257 vo b) x = 238+123. 2+124.1+123.712 3. h +12.5.2 6 ie - 1123.7333 = C) x = 12.4.3 +12 3.74 12 3.6+124.2*124.5+129.0 6 124.8833