The probability that a lab specimen contains high levels of contamination is 0.1
ID: 3306093 • Letter: T
Question
The probability that a lab specimen contains high levels of contamination is 0.10. A group of 3 independent samples are checked.
Round your answers to four decimal places (e.g. 98.7654).
(a) What is the probability that none contain high levels of contamination?
(b) What is the probability that exactly one contains high levels of contamination?
(c) What is the probability that at least one contains high levels of contamination?
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Suppose X has a continuous uniform distribution over the interval [1.2, 5.1].
(a) Determine the mean of X.
(b) Determine the variance of X.
(c) What is P(X < 2.7)?
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In Applied Life Data Analysis (Wiley, 1982), Wayne Nelson presents the breakdown time of an insulating fluid between electrodes at 34 kV. The times, in minutes, are as follows: 0.34, 0.64, 0.82, 1.26, 2.82, 3.21, 4.27, 4.82, 4.90, 6.35, 7.30, 7.90, 8.25, 11.97, 31.68, 32.59, 33.84, 36.78, and 72.96. Calculate the sample mean and sample standard deviation. Round the answers to 3 decimal places.
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An article in Human Factors (June 1989) presented data on visual accommodation (a function of eye movement) when recognizing a speckle pattern on a high-resolution CRT screen. The data are as follows:
36.45, 67.90, 38.77, 42.18, 26.72, 50.77, 39.1, and 50.07.
Calculate the sample mean and sample standard deviation.
Round your answers to 2 decimal places.
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Consider the following two samples:
Sample 1: 11, 8, 7, 6, 7, 5, 11, 5
Sample 2: 11, 5, 11, 5, 7, 11, 7, 5
(a) Calculate the sample range for both samples. Would you conclude that both samples exhibit the same variability?
Sample 1 Range =
Sample 2 Range =
(b) Calculate to 3 decimal places the sample standard deviations for both samples. Do these quantities indicate that both samples have the same variability?
Explanation / Answer
1 a) 1-0.10= 0.90
0.90^3=0.729
b)0.90^2*0.10 =0.00081
C) 1-0.90^3 =1-0.729 =0.271
2 a)mean =[1.2+5.1]/2= 3.15
b) varience =2.67
C) P ( X<2.7 )=P ( (X)/<2.73.15 )
=P (X)/<(2.73.15)/1.632)
Since (x)/=Z and( 2.73.15)/1.632=0.28 we have:
P (X<2.7)=P (Z<0.28)
P (Z<0.28)=0.3897
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