A continuous random variable with cumulative distribution function F has the median value m such that F(m) = 0.5. That is, a random variable is just as likely to be larger than its median as it is to be smaller. A continuous random variable with density f has the mode value x for which f(x) attains its maximum. For each of the following three random variables, (i) state and graph the density function and then (ii) compute the median, mode and mean and show these values on the graph. a) W which is uniformly distributed over the interval [a, b], for some value a, b E R. # Your Code Here YOUR ANSWER HERE b) X which is normal with parameters u and o?, for some value u, o? e R.

Q: 2p=18

A:

Q: Conv or dive?

A:

Q: 64y2+7=32

A: 64y2+7=32

Q: 9w2=12w+5

A:

Q: k2-35=2k

A:

Q: 5+250

A:

Q: 32+4 to Le

A:

Q: 1+2x=77

A:

Q: 1 ER=1 (- 100 k%3D1 2k+1

A:

Q: 2. K 5x+5 16 35

A:

Q: 14 +2

A: Adding Fractions Step 1: Make sure the bottom numbers (the denominators) are the same. Step 2: Add…

Q: 23-8K

A:

Q: Todt 5(-37+4)

A: Given,   ∫  -3t+410dt

Q: 1 +1

A: Given  1+1=

Q: +100 3

A: Our objective is to find the solution of the above expression.

Q: 3=6

A: Suppose the empty box has value x than do the calculation.

Q: 28+76

A:

Q: 16+29

A: Given :16+29

Q: -4d = -64

A: Given  -4d=-64

Q: 11+2x=31

A:

Q: 12

A: Explanation and solutions is given below....

Q: -5 -2 1

A:

Q: 4т V17 3

A: Modulus function always gives positive quantity in result.

Q: 7k = 91

A: We will find the value of k in the following step.

Q: 2t2 – 100 = 0 %3D

A: The solution is, t = ±5√2  The detailed solution is as follows below:

Q: 6+197

A: Given,          6+197

Q: Answer 2

A:

Q: Find L s²-8

A:

Q: 5xn=25

A: 5×n = 25 To find value of n

Q: 16+3•5

A: Given:  16+3*5

Q: カメ Z

A: Find value of  2+2x4

Q: Q7 help me

A:

Q: 2+36=38

A: Given 2+ 36 = 38  Yes the given statement is true .

Q: dx 4е2я — 8ет + 5

A: This question can be solved using the concept of special integrals.

Q: 1. (3, x) 2. (3x + 3, y) 3. (y, cos x) 4. (x + y, x – Y)

A: using online vector field graphing we get 1-A 2-C

Q: -4c +2 ≥10

A: The given inequality is: -4c+2≥10

Q: z? - 34z + 818 = 0

A:

Q: N+2=4

A: The value is:-

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.