Once an individual has been infected with a certain disease, let X represent the time (days) that elapses before the individual becomes infectious. An article proposes a Weibull distribution with a = 2.6, ß= 1.7, and y = 0.5. [Hint: The two-parameter Weibull distribution can be generalized by introducing a third parameter y, called a threshold or location parameter: replace x in the equation below, ²²-1-(x/B)a f(x; a, B) = Ba" 0 x20 x < 0 by xy and x 20 by x 2 y.] (a) Calculate P(1 < X < 2). (Round your answer to four decimal places.) 0.4737 (b) Calculate P(X> 1.5). (Round your answer to four decimal places.) 0.7775 (c) What is the 90th percentile of the distribution? (Round your answer to three decimal places.) 2.843 ✔ days (d) What are the mean and standard deviation of X? (Round your answers to three decimal places.) 1.496 X days mean

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter2: Exponential, Logarithmic, And Trigonometric Functions
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I need help finding the mean for part D, I tried 1.510 and got it wrong and I also tried 1.496 and I also got that wrong. 

Once an individual has been infected with a certain disease, let X represent the time (days) that elapses before the individual becomes infectious. An article proposes a Weibull distribution with a = 2.6, ß = 1.7, and
y = 0.5. [Hint: The two-parameter Weibull distribution can be generalized by introducing a third parameter y, called a threshold or location parameter: replace x in the equation below,
-¹e-(x/B) a
f(x; a, B) =
α
ва
0
Χ20
by x - y and x ≥ 0 by x ≥ y.]
(a) Calculate P(1 < X < 2). (Round your answer to four decimal places.)
0.4737
standard deviation
x < 0
(b) Calculate P(X> 1.5). (Round your answer to four decimal places.)
0.7775
(c) What is the 90th percentile of the distribution? (Round your answer to three decimal places.)
2.843
days
(d) What are the mean and standard deviation of X? (Round your answers to three decimal places.)
mean
X days
✔ days
1.496
0.624
Transcribed Image Text:Once an individual has been infected with a certain disease, let X represent the time (days) that elapses before the individual becomes infectious. An article proposes a Weibull distribution with a = 2.6, ß = 1.7, and y = 0.5. [Hint: The two-parameter Weibull distribution can be generalized by introducing a third parameter y, called a threshold or location parameter: replace x in the equation below, -¹e-(x/B) a f(x; a, B) = α ва 0 Χ20 by x - y and x ≥ 0 by x ≥ y.] (a) Calculate P(1 < X < 2). (Round your answer to four decimal places.) 0.4737 standard deviation x < 0 (b) Calculate P(X> 1.5). (Round your answer to four decimal places.) 0.7775 (c) What is the 90th percentile of the distribution? (Round your answer to three decimal places.) 2.843 days (d) What are the mean and standard deviation of X? (Round your answers to three decimal places.) mean X days ✔ days 1.496 0.624
Once an individual has been infected with a certain disease, let X represent the time (days) that elapses before the individual becomes infectious. An article proposes a Weibull distribution with a = 2.6, ß = 1.7,
and y = 0.5. [Hint: The two-parameter Weibull distribution can be generalized by introducing a third parameter y, called a threshold or location parameter: replace x in the equation below,
_xa – le-(x/B)@
f(x; a, B) = Ba
0
Χ ΣΟ
x < 0
by x - y and x ≥ 0 by x ≥ y.]
(a) Calculate P(1 < X < 2). (Round your answer to four decimal places.)
0.4737✔
(b) Calculate P(X> 1.5). (Round your answer to four decimal places.)
0.7775✔
standard deviation
(c) What is the 90th percentile of the distribution? (Round your answer to three decimal places.)
2.843 days
(d) What are the mean and standard deviation of X? (Round your answers to three decimal places.)
mean
1.510
days
0.624✔
days
Transcribed Image Text:Once an individual has been infected with a certain disease, let X represent the time (days) that elapses before the individual becomes infectious. An article proposes a Weibull distribution with a = 2.6, ß = 1.7, and y = 0.5. [Hint: The two-parameter Weibull distribution can be generalized by introducing a third parameter y, called a threshold or location parameter: replace x in the equation below, _xa – le-(x/B)@ f(x; a, B) = Ba 0 Χ ΣΟ x < 0 by x - y and x ≥ 0 by x ≥ y.] (a) Calculate P(1 < X < 2). (Round your answer to four decimal places.) 0.4737✔ (b) Calculate P(X> 1.5). (Round your answer to four decimal places.) 0.7775✔ standard deviation (c) What is the 90th percentile of the distribution? (Round your answer to three decimal places.) 2.843 days (d) What are the mean and standard deviation of X? (Round your answers to three decimal places.) mean 1.510 days 0.624✔ days
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