Answer:
a) 0.96
b) 0.016
c) 0.018
d) 0.982
e) x = 2
Step-by-step explanation:
We are given with the Probability density function f(x)= 2/x^3 where x > 1.
<em>Firstly we will calculate the general probability that of P(a < X < b) </em>
P(a < X < b) = =
= { Because }
= =
= =
a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }
Comparing with general probability we get,
P(1 < X < 5) = = = 0.96 .
b) P(X > 8) = P(8 < X < ∞) = 1/ - 1/∞ = 1/64 - 0 = 0.016
c) P(6 < X < 10) = = = 0.018 .
d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)
= + (1/ - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982
e) We have to find x such that P(X < x) = 0.75 ;
⇒ P(1 < X < x) = 0.75
⇒ = 0.75
⇒ = 1 - 0.75 = 0.25
⇒ = ⇒ = 4 ⇒ x =
Therefore, value of x such that P(X < x) = 0.75 is 2.
Answer:
9(2p-18) and 2(9p-18)
Step-by-step explanation:
The answer is A or it should be
Answer:
11 ;
Step-by-step explanation:
Given the data:
30 27 22 25 24 25 24 15 35 35 33 52 49 10 27 18 20 23 24 25 30 24 24 24 18 20 25 27 24 32 13 13 21 2 37 35 32 33 29 3 28 28 25 29 31
Number of classes = 5
Class width : Range / number of classes
Class width = (maximum - minimum) / 5
Class width = (52 - 2) / 5 = 50/5 = 10 + 1 = 11
For the frequency table showing class limits, class boundaries, midpoints, frequencies, relative frequencies, and cumulative frequencies and histogram
Kindly check attached picture