Respuesta :
Number of trials (n): 200
Probability of success (p): 1/4
We know that:
[tex]p+q=1\Rightarrow q=\frac{3}{4}[/tex]The distribution is B(200, 1/4), but since the number of trials is big, we can make the normal approximation:
[tex]N(np,npq)=N(200\cdot\frac{1}{4},200\cdot\frac{1}{4}\cdot\frac{3}{4})=N(50,37.5)[/tex]Now, we need to calculate the probability that between 50 and 60 (inclusive) successes occur. That is, our probability of interest is:
[tex]P(50\leq X\leq60)[/tex]Using the normal distribution, we standardize using the z-score formula:
[tex]Z=\frac{x-\mu}{\sigma}[/tex]Where μ is the mean and σ is the standard deviation. Then:
[tex]\begin{gathered} Z_1=\frac{50-50}{\sqrt[]{37.5}}=0 \\ Z_2=\frac{60-50}{\sqrt[]{37.5}}=1.63299 \end{gathered}[/tex]Now, the probability becomes:
[tex]P(50\leq X\leq60)=P(0\leq Z\leq1.63299)[/tex]Using known values on tables:
[tex]P(50\leq X\leq60)=0.4485[/tex]