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Question

Let x be a binomial random variable with n=100 and p=.2. Find approximations to these probabilities: a. P(x$\gt22)b.P(x22) b. P(x\geq22)c.P(2022) c. P(20\ltxx\lt25)d.P(x25) d. P(x\leq$25)

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Given:

n=Sample size=100p=Probability of success=0.2\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: np5np\geq 5 and nq5nq\geq 5.

np=100(0.2)=205nq=n(1p)=100(10.2)=805\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 npnp and divided by the standard deviation npq=np(1p)\sqrt{npq}=\sqrt{np(1-p)}.

z=xnpnp(1p)=22.5100(0.2)100(0.2)(10.2)0.63z=\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)=1P(z<0.63)=10.7357=0.2643P(x>22)=P(x>22.5)=P(z>0.63)=1-P(z<0.63)=1-0.7357=0.2643

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