Math

QuestionQ15: For XB(n,p)X \sim B(n, p) we have Var(X)=0.8E(X)\operatorname{Var}(X)=0.8 E(X). Find pp

Studdy Solution

STEP 1

1. XB(n,p) X \sim B(n, p) denotes a binomial distribution with parameters n n (number of trials) and p p (probability of success).
2. The variance of a binomial distribution is given by Var(X)=np(1p)\operatorname{Var}(X) = np(1-p).
3. The expected value of a binomial distribution is given by E(X)=npE(X) = np.
4. We are given that Var(X)=0.8E(X)\operatorname{Var}(X) = 0.8 E(X).

STEP 2

1. Express the variance and expected value in terms of n n and p p .
2. Set up the equation based on the given relationship between variance and expected value.
3. Solve the equation for p p .

STEP 3

Express the variance and expected value in terms of n n and p p :
Var(X)=np(1p)\operatorname{Var}(X) = np(1-p)
E(X)=npE(X) = np

STEP 4

Set up the equation based on the given relationship Var(X)=0.8E(X)\operatorname{Var}(X) = 0.8 E(X):
np(1p)=0.8(np)np(1-p) = 0.8(np)

STEP 5

Solve the equation for p p :
First, divide both sides by np np (assuming np0 np \neq 0 ):
1p=0.81 - p = 0.8
Subtract 0.8 from both sides:
10.8=p1 - 0.8 = p
p=0.2p = 0.2
The value of p p is:
0.2\boxed{0.2}

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