# Np>5 and nq>5, estimate P(at least 11) with n=13 and p = 0.6 by using the normal distributions as an approximation to the binomial distribution. if np<5 or nq<5 then state the normal approximation is not suitable.p (at least 11) = ?

Np>5 and nq>5, estimate P(at least 11) with n=13 and p = 0.6 by using the normal distributions as an approximation to the binomial distribution. if $np<5\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}nq<5$ then state the normal approximation is not suitable.
p (at least 11) = ?

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Step 1
Given n=12 and p=0.6.
Then,
Mean = np
$=13×0.6$
=7.8
Standard deviation $=\sqrt{np\left(1-p\right)}$
$=\sqrt{7.8\left(1-0.6\right)}$
$=\sqrt{3.12}$
=1.7663
Step 2
So,
$P\left(X\ge 11\right)=P\left(\frac{X-mean}{s\mathrm{tan}darddeviation}\ge \frac{11-mean}{s\mathrm{tan}darddeviation}\right)$
$=P\left(Z\ge \frac{11-7.8}{1.7663}\right)$
$=P\left(Z\ge 1.81\right)$ From the right tailed z - table.
=0.4649
Therefore, p (at least 11)=0.4649