The question provides that:
[tex]\begin{gathered} n=13 \\ p=0.5 \end{gathered}[/tex]
Therefore, we have that:
[tex]q=1-p=1-0.5=0.5[/tex]
To check if we can use the normal distribution as an approximation, we will check the values of np and nq:
[tex]\begin{gathered} np=13\times0.5=6.5 \\ nq=13\times0.5=6.5 \end{gathered}[/tex]
Since,
[tex]\begin{gathered} np\ge5 \\ \text{and} \\ nq\ge5 \end{gathered}[/tex]
then we can use the normal distribution as an approximation.
To evaluate P (at least 10), we are evaluating:
[tex]P(X\ge10)[/tex]
The standard deviation of the distribution is gotten to be:
[tex]\sigma=\sqrt[]{np}=\sqrt[]{6.5}=2.550[/tex]
The mean is 6.5.
Therefore, the Z-score is gotten to be:
[tex]Z=\frac{x-\bar{x}}{\sigma}[/tex]
Hence, it is calculated to be:
[tex]Z=\frac{10-6.5}{2.550}=1.37[/tex]
The probability is therefore given to be:
[tex]P(Z\ge1.37)=Pr(0\le Z)-Pr(0\le Z\le1.37)[/tex]
Using the Probability Distribution Table, we have:
[tex]P(Z\ge1.37)=0.5-0.4147=0.0853\approx0.085[/tex]
Therefore, the answer is:
[tex]P(at\text{ }least\text{ }10)=0.085[/tex]
