Five treatments used during the manufacture of cell phones are being evaluated b
ID: 3178154 • Letter: F
Question
Five treatments used during the manufacture of cell phones are being evaluated by a researcher employed by a manufacturer of phones. Twenty-five batteries from each treatment are tested and the number of days they last beyond 700 days is recorded. Any battery that lasts 60 days beyond 700 is considered a success. The data and some analyses from JMP are provided. Use this information to answer the following questions.
Part A. Suppose the researcher goes to a supplier and randomly chooses one battery from Treatment 1, one from Treatment 2, and one battery from Treatment 3; and then plans to test the three batteries. Use the provided data to estimate the probability that at least 2 of the batteries are considered a success (last beyond 760 days) in this planned test.
--------------------------------------------------------------------------------------------------------------------
PROVIDED DATA:
Treatment 1: 68% successful, 32% unsuccessful
Treatment 2: 56% successful, 44% unsuccessful
Treatment 3: 60% successful, 40% unsuccessful
Treatment 4: 48% successful, 52% unsuccessful
Treatment 5: 4% successful, 96% unsuccessful
-------------------------------------------------------------------------
Treatment 1
Mean: 63.36
Standard Deviation: 14.53983
Standard Error Mean: 2.9079661
Upper 95% Mean: 69.361747
Lower 95% Mean: 57.358253
N = 25
--------------------------------------------------------
Treatment 2
Mean: 63.56
Standard Deviation: 16.452153
Standard Error Mean: 3.2904306
Upper 95% Mean: 70.351115
Lower 95% Mean: 56.768885
N = 25
--------------------------------------------------------
Treatment 3
Mean: 64.8
Standard Deviation: 15.652476
Standard Error Mean: 3.1304952
Upper 95% Mean: 71.261024
Lower 95% Mean: 58.338976
N = 25
--------------------------------------------------------
Treatment 4
Mean: 56.76
Standard Deviation: 14.928385
Standard Error Mean: 2.9856769
Upper 95% Mean: 62.922134
Lower 95% Mean: 50.597866
N = 25
--------------------------------------------------------
Treatment 5
Mean: 38.72
Standard Deviation: 12.102066
Standard Error Mean: 2.4204132
Upper 95% Mean: 43.715487
Lower 95% Mean: 33.724513
N = 25
Explanation / Answer
one battery is chosen each from treatment1, treatment2, treatment3.
PROVIDED DATA:
Treatment 1 (A): 68% successful, 32% unsuccessful
Treatment 2 (B): 56% successful, 44% unsuccessful
Treatment 3 (C): 60% successful, 40% unsuccessful
Events A,B,C represent success of treatments 1,2,3 respectievely.
The events A,B,C are independent.
P(A) = 0.68 ; P(A') = 1-0.68 = 0.32
P(B) = 0.56 ; P(B') = 1-0.56 = 0.44
P(C) = 0.60 ; P(C') = 1-0.60 = 0.40
P(atleast 2 success) = P(A'BC) + P(AB'C) + P(ABC') + P(ABC)
= (0.32*0.56*0.60) + (0.68*0.44*0.6) + (0.68*0.56*0.40) + (0.68*0.56*0.60)
=0.66784.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.