Question:
Consider the following exponential probability density function.
f(x) = 1/5e^(−x/5) for x ≥ 0
(a) Write the formula for P(x ≤ x0). (b) Find P(x ≤ 4). (Round your answer to four decimal places.) (c) Find P(x ≥ 5). (Round your answer to four decimal places.) (d) Find P(x ≤ 6). (Round your answer to four decimal places.) (e) Find P(4 ≤ x ≤ 6). (Round your answer to four decimal places.)
Answer:
(a) P(x ≤ x0) = 1 - e^(−x0/5)
(b) P(x ≤ 4) = 0.5506
(c) P(x ≥ 5) = 0.3678
(d) P(x ≤ 6) = 0.6988
(e) P(4 ≤ x ≤ 6) = 0.1482
Step-by-step explanation:
The standard form of the exponential probability density function is given by
f(x) = 1/μe^(−x/μ)
Where μ is the mean, for the given problem μ = 5
(a) Write the formula for P(x ≤ x0)
P(x ≤ x0) = 1 - e^(−x0/5)
(b) Find P(x ≤ 4)
P(x ≤ 4) = 1 - e^(−4/5)
P(x ≤ 4) = 1 - 0.4493
P(x ≤ 4) = 0.5506
(c) Find P(x ≥ 5)
P(x ≥ 5) = e^(−5/5)
P(x ≥ 5) = 0.3678
(d) Find P(x ≤ 6)
P(x ≤ 6) = 1 - e^(−6/5)
P(x ≤ 6) = 1 - 0.3011
P(x ≤ 6) = 0.6988
(e) Find P(4 ≤ x ≤ 6)
P(4 ≤ x ≤ 6) = e^(−4/5) - e^(−6/5)
P(4 ≤ x ≤ 6) = 0.4493 - 0.3011
P(4 ≤ x ≤ 6) = 0.1482
First you would find a common denominator. If 40 is the LCM then you would have to add playing games, instruction,warm-up, and cool-down. Then you will subtract the sum from the total class time and bam! You get your answer!
<u>Answer:</u>
X and Y are stochastically dependent RVs .
<u>Step-by-step explanation:</u>
Let ,
X = sum of the values that come up after throwing n (≥ 1) fare dice.
Y = number of times an odd number come up.
Let, n = 3
then, P(X =6) = p (say) clearly 0 < p < 1
and P (Y = 3) = 
And,
P( X = 6, Y = 3) = 0 ≠ 
Hence, X and Y are stochastically dependent RVs
Answer
Step-by-step explanation:
C
Answer:
C. 4 and 5
Step-by-step explanation:
29/7 = 4 1/7
