# Thread: Estimation,Confidence intervals and tests

1. ## Estimation,Confidence intervals and tests

Can anyone help me to solve the following problem?

The 25 children in a class each roll a fair die 30 times and record the number of sixes they obtain. Find an estimate of the probability that the mean number of sixes recorded for the class is less than 4.5

2. Originally Posted by Sashikala
Can anyone help me to solve the following problem?

The 25 children in a class each roll a fair die 30 times and record the number of sixes they obtain. Find an estimate of the probability that the mean number of sixes recorded for the class is less than 4.5
The expected value or mean number of 6's rolled for each child will be 30/6 = 5 and as there are 25 students the mean number of 6's rolled for the calss will be 750/6 = 125.

Is there any indication of how the data is spread? I.e if the data is normal and has a standard deviation then you can use a z-score

For example (x<4.5) = (Z< (4.5 - mean)/std)

Then we can find a value in a standard normal distribution table.

I have created a simulation for you below, could be helpful!

3. ## Estimation,Confidence intervals and tests

Thanks for the response.
Yes, We need to use the following
P(x<4.5) = P(Z< (4.5 - mean)/std)

Here mean will be 25*(30/6)/25=5
How to get the value for std?

4. Originally Posted by Sashikala
Thanks for the response.
Yes, We need to use the following
P(x<4.5) = P(Z< (4.5 - mean)/std)

Here mean will be 25*(30/6)/25=5
How to get the value for std?
I would say that because we are looking for the probabilty of throwing a 6 then the problem can be looked as a binomial random variable.

For the entire class we have
n = 750,
p (success) = 1/6,
q = 1-p (failure) = 5/6

the standard deviation $\displaystyle \sigma$ can be found as

$\displaystyle \sigma = \sqrt{npq}$

$\displaystyle \sigma = \sqrt{750.\frac{1}{6}.\frac{5}{6}}$

$\displaystyle \sigma = 10.21$

Or per student

n = 30,
p (success) = 1/6,
q = 1-p (failure) = 5/6

the standard deviation $\displaystyle \sigma$ can be found as

$\displaystyle \sigma = \sqrt{npq}$

$\displaystyle \sigma = \sqrt{30.\frac{1}{6}.\frac{5}{6}}$

$\displaystyle \sigma = 2.04$

5. ## Estimation,Confidence intervals and tests

You explained it very clearly.
Oh...., Thank you very much.!!!