Given the random variable X with distribution. x P(X = x) 0 0.15 0.5 0.24 1 0.09 1.5 0.34 2.0 0.18 Calculate the following: P(0 < X ≤ 1.5) If pX (x) is the probability mass function for X, what is pX (1) If FX (x) is the cumulative probability distribution for X, what is FX (1.5)
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Given the random variable X with distribution.
| x | P(X = x) |
| 0 | 0.15 |
| 0.5 | 0.24 |
| 1 | 0.09 |
| 1.5 | 0.34 |
| 2.0 | 0.18 |
Calculate the following:
- P(0 < X ≤ 1.5)
- If pX (x) is the
probability massfunction for X, what is pX (1) - If FX (x) is the cumulative probability distribution for X, what is FX (1.5)
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- Let X be a random variable with the following probability distribution. Value x of X P(X=x) 10 20 30 40 50 0.10 0.30 0.20 0.25 0.15 Complete the following. (If necessary, consult a list of formulas.) (a) Find the expectation E (X) of X. E (X) = 0 (b) Find the variance Var(X) of X. Var(x) =Let X be a random variable with the following probability distribution. Value x of X P(X=x) 2 0.05 3 0.05 4 0.05 5 0.85 Complete the following. (If necessary, consult a list of formulas.) (a) Find the expectation E(X) of X. E (X) = 0 (b) Find the variance Var(X) of X. Var(x) = ☐ Check B X GI. Which of the following are discrete probability distributions? (Yes or No) 1. 3 4 5 P(x) 0.10 0.20 0.25 0.40 0.05 2. 3 4 P(x) 0,05 0.25 0,33 0.28 0,08 3. 4 5 P(x) 0.08 0.25 0.34 0.31 0,04 4. 2 3 4 5 P(x) 0.03 0.22 1.01 0.23 0.02 5. 3 4 5 P(x) 0.05 0.27 0.34 0.28 0.06 6. 2 3 4 5 P(x) 1 3 3 10 10 10 7. 3. 4 P(x) 15 15 8. 2 4 5 6 P(x) 4. 25 25 25 25 9. 3 4 P(x) 1. 3. 20 20 1 10 10 10. 1 3 4 P(x) 0.212 0.113 0.125 0.224 0.306 -
- Suppose X takes the value 1 with probability P, and the value 0 with probability (1-P). The Probability distribution of X is P(X) 1-P 1 P The expected value of X, E(X) is:Let X be a discrete random variable with the following cumulative.distribution function 0. x <0 0.2 0Sx<1 15x<3 3The random variable x has the following discrete probability distribution. 10 0.1 11 12 13 14 (x)d The values that x can assume are mutually exclusive events. Find P(x > 12). 0.2 0.1 0.3 0.3 P(x>12) =D 0 %3D esc F1 000Let X be a random variable with the following probability distribution. Value x of X P(X=x) 3 4 5 5.0 6 0.10 0.05 0.40 0.45 Complete the following. (If necessary, consult a list of formulas.) (a) Find the expectation E (X) of X. E (X) = 0 (b) Find the variance Var(X) of X. Var(x) = 0An unfair die looks like an ordinary six-sided die but the outcomes are not equally likely. The probability distribution of the face value, XX, is as follows: xixi 1 2 3 4 5 6 Total P(X=xi)P(X=xi) 0.15 0.16 0.33 0.2 0.11 0.05 1 The expected value is computed, E[X]=3.11E[X]=3.11. To find the variance and standard deviation the following table was set up: xixi P(X=xi)P(X=xi) (xi−E[X])2(xi-E[X])2 (xi−E[X])2P(X=xi)(xi-E[X])2P(X=xi) 1 0.15 (1−3.11)2=(−2.11)2=4.4521(1-3.11)2=(-2.11)2=4.4521 4.4521⋅0.15=0.66784.4521⋅0.15=0.6678 2 (2−3.11)2=(−1.11)2=1.2321(2-3.11)2=(-1.11)2=1.2321 1.2321⋅0.16=0.19711.2321⋅0.16=0.1971 3 0.33 (3−3.11)2=(−0.11)2=(3-3.11)2=(-0.11)2= 0.0121⋅0.33=0.0040.0121⋅0.33=0.004 4 0.2 (4−3.11)2=(0.89)2=0.7921(4-3.11)2=(0.89)2=0.7921 0.7921⋅0.2=0.7921⋅0.2= 5 (5−3.11)2=(1.89)2=3.5721(5-3.11)2=(1.89)2=3.5721 3.5721⋅0.11=0.39293.5721⋅0.11=0.3929 6 0.05 8.3521⋅0.05=0.41768.3521⋅0.05=0.4176 E[X]=3.11E[X]=3.11 Total:…Let X be a random variable with the following probability distribution.Calculate the following Find the expectation E (X) of XE (X) =? Find the variance Var (X) of X.Var (X) =?Recommended textbooks for youA First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON
A First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON