For a normal distribution of $N(30,10)$ and sagnificence of 5% (α=0.05),

the sample mean is 32.5 and the population size is 64.

H0: $μ=30$

H1: $μ>30$

A. What is the probability for a second type error?(β)

What I did was, $=P( accept H\_0 | H\_0 false)=P(X ̅<30 | μ=32.5)=Φ((30-32.5)/(10/√64))=1-Φ(2)=0.0227$

B. What is the sample size needed for a $90$% discovery chance of the second type error?

what I did was,

$n≥((Z\_(1-α)+Z\_(1-β) )^2 σ^2)/(μ\_1-μ\_0 )^2 =((1.65+1.285)^2 ×10^2)/(32.5-30)^2 =137.83$

My problem is,

In part B, the probability for a second type error is $0.1$, and the sample size I get is 138, BUT in part A the probability for a second type error I get is 0.027 and the sample size is 64.

How is that possible that I get a bigger sample size for a higher probability for an error?

The full excercise,
[![excercise](https://i.stack.imgur.com/9U4gT.gif)](https://i.stack.imgur.com/9U4gT.gif)