# Normal/t distributions

1. Jul 11, 2013

### Phox

1. The problem statement, all variables and given/known data

Let X1, X2,...,X16 be a random sample of size 16 from N(μ=50, σ2=100) distribution.

Find P(Xbar > 50 + .6505(s))

2. Relevant equations

Z= (xbar - μ)/(σ/√n)

3. The attempt at a solution

So I know the solution of this problem is given by P( T(15 d.f.) > 2.602 ) but I'm not sure why.

I notice that you get the value 2.602 by plugging (50 + .6505(s)) in for xbar in the Z equation.

I guess I'm not really sure why we use the T distribution here and how you're supposed to come up with the value 2.602. 15 degrees of freedom makes sense.

2. Jul 11, 2013

### Staff: Mentor

You have to use the Student's t distribution because your sample size is so small. As I recall the cutoff is about n = 28. For your problem, n = 16.

3. Jul 11, 2013

### Phox

Ok. That makes sense.

I suppose I'm confused because in the previous problem we were asked to find P(xbar >52) and we used z scores/normal cdf

4. Jul 11, 2013

### Ray Vickson

What, exactly, do you mean by 's' in the expression Xbar > 50 + 0.6505 s ? If 's' is the sample standard deviation, you need to use the t-distribution because you are essentially asking for the distribution of T = (Xbar - μ)/s (or maybe (Xbar - μ)/(s/√n), depending on what s actually represents). You would use the normal distribution if you wanted Xbar > 50 + 0.6505 σ with known σ = 10. In other words, use N for (Xbar - μ)/σ and use T for (Xbar - μ)/s.

5. Jul 11, 2013

### Phox

's' represents the sample standard deviation.

I'm still a bit unclear as to how we come up with the value 2.602

6. Jul 11, 2013

### Ray Vickson

$$2.602 = 4 \times 0.6505 = \sqrt{16} \times 0.6505$$