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# Let x be a binomial random variable with n=100 and p=.2. Find approximations to these probabilities: a. P(x$\gt$22) b. P(x$\geq$22) c. P(20$\lt$x$\lt$25) d. P(x$\leq$25)

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Given:

\begin{align*} n&=\text{Sample size}=100 \\ p&=\text{Probability of success}=0.2 \end{align*}

(a) Requirements for a normal approximation of the binomial distribution: $np\geq 5$ and $nq\geq 5$.

\begin{align*} np&=100(0.2)=20\color{#4257b2}\geq 5 \\ nq&=n(1-p)=100(1-0.2)=80\color{#4257b2}\geq 5 \end{align*}

Thus the requirements are satisfied and we can then approximate the binomial distribution by the normal distribution.

The z-score is the value (using the continuity correction) decreased by the mean $np$ and divided by the standard deviation $\sqrt{npq}=\sqrt{np(1-p)}$.

$z=\dfrac{x-np}{\sqrt{np(1-p)}}=\dfrac{22.5-100(0.2)}{\sqrt{100(0.2)(1-0.2)}}\approx 0.63$

Determine the corresponding normal probability using the normal probability table in the appendix:

$P(x>22)=P(x>22.5)=P(z>0.63)=1-P(z<0.63)=1-0.7357=0.2643$

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