Question

Question:
A random variable X has a probability distribution as follows:
1
2
X
PX=x)
1
|
O 0
3
2k
3k
13k
2k
Then the probability that Pr(X < 2 ) is equal to​

Answer 1

Answer:

Sum of all probabilities must be equal to 1

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1,

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 10

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K=

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 10

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101 The mean is 1.K+2.2K+3.2K+4.3K+5K

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101 The mean is 1.K+2.2K+3.2K+4.3K+5K 2

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101 The mean is 1.K+2.2K+3.2K+4.3K+5K 2 +6.2K

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101 The mean is 1.K+2.2K+3.2K+4.3K+5K 2 +6.2K 2

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101 The mean is 1.K+2.2K+3.2K+4.3K+5K 2 +6.2K 2 +7.7K

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101 The mean is 1.K+2.2K+3.2K+4.3K+5K 2 +6.2K 2 +7.7K 2

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101 The mean is 1.K+2.2K+3.2K+4.3K+5K 2 +6.2K 2 +7.7K 2 +7K

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101 The mean is 1.K+2.2K+3.2K+4.3K+5K 2 +6.2K 2 +7.7K 2 +7K=30K+66K

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101 The mean is 1.K+2.2K+3.2K+4.3K+5K 2 +6.2K 2 +7.7K 2 +7K=30K+66K 2

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101 The mean is 1.K+2.2K+3.2K+4.3K+5K 2 +6.2K 2 +7.7K 2 +7K=30K+66K 2 =3+0.66=3.66

Sum of all probabilities must be equal to 1So we get K+2K+2K+3K+K 2 +2K 2 +7K 2 +K=1⇒10K 2 +9K=1⇒K=−1, 101 But probability cannot be negative , therefore the value of K= 101 The mean is 1.K+2.2K+3.2K+4.3K+5K 2 +6.2K 2 +7.7K 2 +7K=30K+66K 2 =3+0.66=3.66P(0<x<5)=P(x=1)+P(x=2)+P(x=3)+P(x=4)=K+2K+2K+3K=8K=0.8