# Finding a Confidence Interval (No Standard Deviation nor Mean given)

#### hustisyawright

##### New member
Hello, I am having a hard time solving this problem: The interval (20.6, 26.2) is a 90% confidence interval. Construct a 95% confidence interval for the same data. Use n=30. Thank you.

#### Subhotosh Khan

##### Super Moderator
Staff member
Hello, I am having a hard time solving this problem: The interval (20.6, 26.2) is a 90% confidence interval. Construct a 95% confidence interval for the same data. Use n=30. Thank you.
Are the data normally distributed?

From the data given to you:

Can you calculate the mean (assuming normal distribution)?

#### HallsofIvy

##### Elite Member
Since you are given "n= 30" it sounds like you have a binomial distribution with some probability "p". For such a distribution the mean is np= 30p and the standard deviation is $$\displaystyle \sqrt{np(1-p)}$$. Knowing the "90% confidence interval" you should be able to calculate p and then the 95% confidence interval.

#### hustisyawright

##### New member
Since you are given "n= 30" it sounds like you have a binomial distribution with some probability "p". For such a distribution the mean is np= 30p and the standard deviation is $$\displaystyle \sqrt{np(1-p)}$$. Knowing the "90% confidence interval" you should be able to calculate p and then the 95% confidence interval.
Since you are given "n= 30" it sounds like you have a binomial distribution with some probability "p". For such a distribution the mean is np= 30p and the standard deviation is $$\displaystyle \sqrt{np(1-p)}$$. Knowing the "90% confidence interval" you should be able to calculate p and then the 95% confidence interval.

Thank you. This is very helpful.