Q1 (a) A random variable X has the following probability function: Х O 1 2 3 4 5 and P(x) 0 k 2k 2k 3k 5k
1 answer:
Answer:
(i) It is known that the sum of probabilities of a probability distribution of random variables is one.
∴0+k+2k+3k+k
2
+2k
2
+(7k
2
+k)=1
⇒10k
2
+9k−1=0
⇒(10k−1)(k+1)=0
⇒k=−1,
10
1
k=−1 is not possible as the probability of an event is never negative.
∴k=
10
1
(ii) P(X<3)=P(X=0)+P(X=1)+P(X=2)
=0+k+2k
=3k
=3×
10
1
=
10
3
(iii) P(X>6)=P(X=7)
=7k
2
+k
=7×
10
1
2
+
10
1
=
100
7
+
10
1
=
100
17
(iv) P(0<X<3)=P(X=1)+P(X=2)
=k+2k
=3k
=3×
10
1
=
10
3
You might be interested in
The answer to this questuon is 2
Answer:
Alex can drive 480miles in one day period.
Step-by-step explanation:
$162- $42 =$120
$120/0.25¢=480
480miles
The vertex of <span>y = 7x2 + 14x + 4</span> is at
(-1,-3) (see attachment)
2 = 2(-3) + 8
2 = -6 + 8
2 = 2
true.
2 = 3 + 1
2 = 4
false.
this is not a solution because both equations have to be true.
