# Normal Approximation probability question

#### devimon

##### New Member
Consider a four-sided die loaded so that 1 shows 40% of the time. If we roll this die 50,1000,2000,4000,8000 times, find the probability in each case that at least 42% of the rolls are 1's

P = 0.4, n = 50,1000,2000,4000,8000
mean p^ = p = 0.4
standard deviation p^ =

Sq. rt 0.4 ∗ 0.6 /
50 et al

= 0.069282, 0.015492, 0.010954, 0.00774597, 0.0054772 ---> i don't get how to arrive at these numbers
p ≥ 0.42
Z ≥0.42 − 0.40. /069282 et al

Z ≥ 0.29, 1.29, 1.83, 2.58, 3.65
1 − 0.6141, 0 .9015, 0.9664, 0.9951, approx. 1
0.3859, 0.0985, 0.0336, 0.0049, off the chart, practically 0

#### BGM

##### TS Contributor
n = 50,1000,2000,4000,8000
Sq. rt 0.4 ∗ 0.6 /
50 et al
So you just compute something like

$$\sqrt{\frac {0.4 \times 0.6} {n}}$$ for $$n = 50, 1000, 2000, 4000, 8000$$

E.g. in R,
Code:
> n<-c(50,1000,2000,4000,8000)
> sqrt(0.4*0.6/n)
[1] 0.069282032 0.015491933 0.010954451 0.007745967 0.005477226