Answer:
0.4801
Step-by-step explanation:
This is a binomial distribution question.
It can be approximated using normal distribution if the following conditions are met:
np > 10
n(1-p) > 10
Here,
n = 29
p = 0.487
So,
np = 14.12
n(1-p) = 14.88
So, we can use normal approximation here:
Binomial: X ~ B(n,p) becomes
Normal Approx: X~ N([tex]np,\sqrt{np(1-p)}[/tex])
Mean is:
[tex]\mu=np=14.123[/tex]
Standard Deviation is:
[tex]\sigma=\sqrt{np(1-p)} =2.69[/tex]
We need probability of less than or equal to 14, so we can say:
P(x ≤ 14)
Using [tex]z=\frac{x-\mu}{\sigma}[/tex], we have:
P(x ≤ 14) = [tex]P(\frac{x-\mu}{\sigma} \leq \frac{14-14.123}{2.69})\\=P(z \leq -0.05)\\=0.4801[/tex]
Note: We used z table in the last line
So the probability is 0.4801
