Question

I need part b and d, which the answers are not 1.0733 and 0.3394. Thanks.

Part II: Binomial Distribution & the Central Limit Theorem a) Imagine sampling 5 values from Bin(15,.2), a binomial distribution with 15 trials and success probability of .2 What should be the expected value (mean) of this sample? You should use the CLT (central limit theorem) and should not need to do any coding to answer this. Answer: 33 Part lI: Binomial Distribution & the Central Limit Theorem b) Imagine sampling 5 values from Bin(15,.2), a binomial distribution with 15 trials and success probability of .2 What should be the variance of this sample? Please round to 4 decimal places. You should use the CLT (central limit theorem) and should not need to do any coding to answer this. Answer: 1.0733| Check

Answer 1

b) Here we have binomial experiment with k= 15, p = 0.2.

So variance of binomial distribution is :

v( x) = k * p * (1 - p ) = 15 * 0.2 * 0.8 = 2.4

According top CLT , mean of sample mean is  $\mu _{\overline{x}}=\mu$ and standard deviation is $\sigma _{\overline{x}}=\sigma/\sqrt{n}$

Hence variance of sample mean is $\sigma _{\overline{x}}^{2}=\sigma^{2}/n$

Here 5 samples are drawn from Binomial distribution. So n = 5

Hence variance of sample is,

variance = 2.4/ 5 = 0.4800

d)  

Here we have binomial experiment with k= 15, p = 0.2.

So variance of binomial distribution is :

v( x) = k * p * (1 - p ) = 15 * 0.2 * 0.8 = 2.4

According top CLT , mean of sample mean is  $\mu _{\overline{x}}=\mu$ and standard deviation is $\sigma _{\overline{x}}=\sigma/\sqrt{n}$

Hence variance of sample mean is $\sigma _{\overline{x}}^{2}=\sigma^{2}/n$

Here 50 samples are drawn from Binomial distribution. So n = 50

Hence variance of sample is,

variance = 2.4/ 50 = 0.0480