Answer to Question #294789 in Statistics and Probability for John

Question #294789

A random sample of size 50 is drawn from binomial distribution with parameters n=100 and p=0.3. what is the probability that the sample mean is



a)larger than 3.9



b) between 4.1 and 4.4



c) smaller than 4.0

1
Expert's answer
2022-02-17T03:34:00-0500

Let XX be a binomial rv based on nn trials with success probability p.p. Then if the

binomial probability histogram is not too skewed, XX has approximately a

normal distribution with μ=np\mu=np and σ=npq.\sigma=\sqrt{npq}.

In practice, the approximation is adequate provided that both np10np\ge 10 and nq10.nq\ge 10.

Given n=100,p=0.3,q=1p=10.3=0.7.n=100, p=0.3, q=1-p=1-0.3=0.7.

np=100(0.3)=3010np=100(0.3)=30\ge 10

nq=100(0.7)=7010nq=100(0.7)=70\ge 10

We can use normal approximation for binomial distribution with μ=np=100(0.3)=30,σ=npq=100(0.3)(0.7)=21\mu=np=100(0.3)=30, \sigma=\sqrt{npq}=\sqrt{100(0.3)(0.7)}=\sqrt{21}

Let Xˉ=\bar{X}= the sample mean: XˉN(μ,σ2/n1)\bar{X}\sim N(\mu, \sigma^2/n_1)

Given n1=50.n_1=50.

a)


P(Xˉ>3.9)=1P(Z3.93021/50)P(\bar{X}>3.9)=1-P(Z\le\dfrac{3.9-30}{\sqrt{21}/\sqrt{50}})

1P(Z40.2732)1\approx1-P(Z\le-40.2732)\approx1


b)


P(4.1<Xˉ<4.4)=P(Z<4.43021/50)P(4.1<\bar{X}<4.4)=P(Z<\dfrac{4.4-30}{\sqrt{21}/\sqrt{50}})

P(Z4.13021/50)-P(Z\le\dfrac{4.1-30}{\sqrt{21}/\sqrt{50}})




P(Z<39.5017)P(Z39.9646)0\approx P(Z<-39.5017)-P(Z\le-39.9646)\approx0


c)


P(Xˉ<4.0)=P(Z<4.03021/50)P(\bar{X}<4.0)=P(Z<\dfrac{4.0-30}{\sqrt{21}/\sqrt{50}})

P(Z<40.1189)0\approx P(Z<-40.1189)\approx0


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment