Sample Distribution Quiz1 - No20

Today I am going to solve a simple quiz based on the knowledge we've learned during last few post.

My quiz is like this.

Can you figure out the probability that sample mean which sampled from the normal distribution (MEAN = 10, SD = 10)  is over 11 ?   (Sample number = 50) 

I emphasized before that we can predict population mean from the sample distribution. however, this is an opposite one because we already know the population mean and standard deviation.

Let's take a look at this step by step.

According to the central limit theorem, if you take a sample over and over again with a sample size is 50 from the normal distribution, then your sample distribution will be approximately close to the normal distribution ( mean = 10 and standard deviation of sample distribution = 10 / sqrt (50) )

Now, you can find out that cumulative probability of 11 by using R command.

> pnorm(11,10, 10/sqrt(50))

[1] 0.7602499

But our question is probability that average is over 11 so you have to extract above cumulative probability from 1 (total probability)

As a result , answer is 23 percent.

> 1- pnorm(11,10, 10/sqrt(50))

[1] 0.239750

* I've tested with a real data a couple of times and  I've found out that  approximately 23% chance was appeared. take a look at the below. it's quite a interesting.

> mean(rnorm(50,10,10))

[1] 11.48688

> mean(rnorm(50,10,10))

[1] 9.972127

> mean(rnorm(50,10,10))

[1] 9.825054

> mean(rnorm(50,10,10))

[1] 9.742521

* I think you can quess that probability that average is over 13 could be lower than 11.

 let's simulate is with other parameter.

As you can see, it's only 1.6 %, it's quite low chance.

> 1- pnorm(13,10, 10/sqrt(50))

[1] 0.01694743

it's interesting.