Answer: 0.111
Step-by-step explanation:
Given: The probability of winning on an arcade game : p=0.612
The number of times you play the arcade game : n=21
We assume this is a normal distribution , then we have
[tex]\text{Mean}=\mu=np=21(0.612)=12.852[/tex]
[tex]\text{Standard deviation}=\sigma=\sqrt{np(1-p)}\\\\=\sqrt{21(0.612)(1-0.612)}\approx2.23[/tex]
Let X be a binomial variable .
Then , the z score for x= 10 will be :-
[tex]z=\dfrac{x-\mu}{\sigma}=\dfrac{10-12.852}{2.33}\approx-1.22[/tex]
Now, the probability of winning no more than 10 times :_
[tex]P(x\leq10)=P(\leq-1.22)=0.1112324\approx0.111[/tex]
Hence, the probability of winning no more than 10 times =0.111
