# 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) = ?

Normal distributions
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 $$\displaystyle{n}{p}{<}{5}{\quad\text{or}\quad}{n}{q}{<}{5}$$</span> then state the normal approximation is not suitable.
p (at least 11) = ?

2021-01-28
Step 1
Given n=12 and p=0.6.
Then,
Mean = np
$$\displaystyle={13}\times{0.6}$$
=7.8
Standard deviation $$\displaystyle=\sqrt{{{n}{p}{\left({1}-{p}\right)}}}$$
$$\displaystyle=\sqrt{{{7.8}{\left({1}-{0.6}\right)}}}$$
$$\displaystyle=\sqrt{{{3.12}}}$$
=1.7663
Step 2
So,
$$\displaystyle{P}{\left({X}\ge{11}\right)}={P}{\left(\frac{{{X}-{m}{e}{a}{n}}}{{{s}{\tan{{d}}}{a}{r}{d}{d}{e}{v}{i}{a}{t}{i}{o}{n}}}\ge\frac{{{11}-{m}{e}{a}{n}}}{{{s}{\tan{{d}}}{a}{r}{d}{d}{e}{v}{i}{a}{t}{i}{o}{n}}}\right)}$$
$$\displaystyle={P}{\left({Z}\ge\frac{{{11}-{7.8}}}{{{1.7663}}}\right)}$$
$$\displaystyle={P}{\left({Z}\ge{1.81}\right)}$$ From the right tailed z - table.
=0.4649
Therefore, p (at least 11)=0.4649