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Find the probability of 3 success for the binomial experiment with 7 trial and the success probability of 0.3. Then find the mean and standard deviation. Write the formula substitute

the values.

User IanB
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1 Answer

5 votes

Answer:


P(x=3)=0.2269

Mean=2.1

Standard deviation=1.21

Explanation:

We are given that

n=7

Probability of success, p=0.3

q=1-p=1-0.3=0.7

We have to find the probability of 3 success for the binomial experiment and find the mean and standard deviation.

Binomial distribution formula


P(X=x)=nC_xp^(x)q^(n-x)

Using the formula


P(x=3)=7C_3(0.3)^3(0.7)^(7-3)


P(x=3)=7C_3(0.3)^3(0.7)^(4)


P(x=3)=(7!)/(3!4!)(0.3)^3(0.7)^(4)


P(x=3)=(7* 6* 5* 4!)/(3* 2* 1* 4!)(0.3)^(3)(0.7)^(4)

Using the formula


nC_r=(n!)/(r!(n-r)!)


P(x=3)=0.2269

Now,

Mean,
\mu=np=7* 0.3=2.1

Standard deviation,
\sigma=√(np(1-p))

Standard deviation,
\sigma=√(7* 0.3* 0.7)

Standard deviation,
\sigma=1.21

User Mafii
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