EBK APPLIED CALCULUS, ENHANCED ETEXT
6th Edition
ISBN: 9781119399353
Author: DA
Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT
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Chapter 7.2, Problem 17P
To determine
(a)
To graph:
The cumulative distribution function for the shelf life of bananas.
To determine
(b)
The probability that banana lasts between 1 and 2 weeks.
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Students have asked these similar questions
With the known values of a = 200,000 and b = 230,000, the first equation for the probability density function for the sales price of a home is found as follows.
f(x) = 1/b – a, a ≤ x ≤ b
= 1/230,000 − 200000, 200,000 ≤ x ≤ 230,000
= 1/30000, 200,000 ≤ x ≤ 230,000
Everywhere else, the probability density function will just be 0. Therefore, the full probability density function for the sales price of a home follows.
f(x) = _______________ 200,000 ≤ x ≤ 230,000
elsewhere
Could you show me step by step how to find the density function?
•The answer should be f(x)=1/2u•exp^-x/(2u)
It should be an exponential distribution, but why?
The bar chart above is to be used for problems 17-19. It represents the probability density function for x, where x can take on the values of 1, 2, 3 and so on up to 10.
This probability density function is:
a.
discrete
b.
continuous
c.
normal
d.
transverse
Chapter 7 Solutions
EBK APPLIED CALCULUS, ENHANCED ETEXT
Ch. 7.1 - Prob. 1PCh. 7.1 - Prob. 2PCh. 7.1 - Prob. 3PCh. 7.1 - Prob. 4PCh. 7.1 - Prob. 5PCh. 7.1 - Prob. 6PCh. 7.1 - Prob. 7PCh. 7.1 - Prob. 8PCh. 7.1 - Prob. 9PCh. 7.1 - Prob. 10P
Ch. 7.1 - Prob. 11PCh. 7.1 - Prob. 12PCh. 7.1 - Prob. 13PCh. 7.1 - Prob. 14PCh. 7.1 - Prob. 15PCh. 7.1 - Prob. 16PCh. 7.1 - Prob. 17PCh. 7.1 - Prob. 18PCh. 7.2 - Prob. 1PCh. 7.2 - Prob. 2PCh. 7.2 - Prob. 3PCh. 7.2 - Prob. 4PCh. 7.2 - Prob. 5PCh. 7.2 - Prob. 6PCh. 7.2 - Prob. 7PCh. 7.2 - Prob. 8PCh. 7.2 - Prob. 9PCh. 7.2 - Prob. 10PCh. 7.2 - Prob. 11PCh. 7.2 - Prob. 12PCh. 7.2 - Prob. 13PCh. 7.2 - Prob. 14PCh. 7.2 - Prob. 15PCh. 7.2 - Prob. 16PCh. 7.2 - Prob. 17PCh. 7.2 - Prob. 18PCh. 7.2 - Prob. 19PCh. 7.2 - Prob. 20PCh. 7.2 - Prob. 21PCh. 7.3 - Prob. 1PCh. 7.3 - Prob. 2PCh. 7.3 - Prob. 3PCh. 7.3 - Prob. 4PCh. 7.3 - Prob. 5PCh. 7.3 - Prob. 6PCh. 7.3 - Prob. 7PCh. 7.3 - Prob. 8PCh. 7.3 - Prob. 9PCh. 7.3 - Prob. 10PCh. 7.3 - Prob. 11PCh. 7.3 - Prob. 12PCh. 7 - Prob. 1SYUCh. 7 - Prob. 2SYUCh. 7 - Prob. 3SYUCh. 7 - Prob. 4SYUCh. 7 - Prob. 5SYUCh. 7 - Prob. 6SYUCh. 7 - Prob. 7SYUCh. 7 - Prob. 8SYUCh. 7 - Prob. 9SYUCh. 7 - Prob. 10SYUCh. 7 - Prob. 11SYUCh. 7 - Prob. 12SYUCh. 7 - Prob. 13SYUCh. 7 - Prob. 14SYUCh. 7 - Prob. 15SYUCh. 7 - Prob. 16SYUCh. 7 - Prob. 17SYUCh. 7 - Prob. 18SYUCh. 7 - Prob. 19SYUCh. 7 - Prob. 20SYUCh. 7 - Prob. 21SYUCh. 7 - Prob. 22SYUCh. 7 - Prob. 23SYUCh. 7 - Prob. 24SYUCh. 7 - Prob. 25SYUCh. 7 - Prob. 26SYUCh. 7 - Prob. 27SYUCh. 7 - Prob. 28SYUCh. 7 - Prob. 29SYUCh. 7 - Prob. 30SYU
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- Question 1 : Suppose that the probability density function (p.d.f.) of the life (in weeks) of a certain part is f(x) = 3 x 2 (400)3 , 0 ≤ x < 400. (a) Compute the probability the a certain part will fail in less than 200 weeks. (b) Compute the mean lifetime of a part and the standard deviation of the lifetime of a part. (c) To decrease the probability in part (a), four independent parts are placed in parallel. So all must fail, if the system fails. Let Y = max{X1, X2, X3, X4} denote the lifetime of such a system, where Xi denotes the lifetime of the ith component. Show that fY (y) = 12 y 11 (400)12 , y > 0. Hint : First construct FY (y) = P(Y ≤ y), by noticing that {Y ≤ y} = {X1 ≤ y} ∩ {X2 ≤ y} ∩ {X3 ≤ y} ∩ {X4 ≤ y}. (d) Determine P(Y ≤ 200) and compare it to the answer in part (a)arrow_forwardThe probability density function of X, the lifetime of a certain device in months, is given by f(x)=6/(x^2) if x > 6 and f(x) = 0 if x < 6. Find P(X > 50). Also, what is the cumulative distribution function of X on x > 6?arrow_forwardIf the geometric mean of two observations is (1/n) times their harmonic mean, prove that the rate of the two observations is: ( 1+✓1n²) : ( 1. ✓1- n²)?arrow_forward
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